I saw some usage of (void*) in printf().
If I want to print a variable's address, can I do it like this:
int a = 19;
printf("%d", &a);
I think, &a is a's address which is just an integer, right?
Many articles I read use something like this:
printf("%p", (void*)&a);
What does %p stand for? (A pointer?)
Why use (void*)? Can't I use (int)&a instead?
Pointers are not numbers. They are often internally represented that way, but they are conceptually distinct.
void* is designed to be a generic pointer type. Any pointer value (other than a function pointer) may be converted to void* and back again without loss of information. This typically means that void* is at least as big as other pointer types.
printfs "%p" format requires an argument of type void*. That's why an int* should be cast to void* in that context. (There's no implicit conversion because it's a variadic function; there's no declared parameter, so the compiler doesn't know what to convert it to.)
Sloppy practices like printing pointers with "%d", or passing an int* to printf with a "%p" format, are things that you can probably get away with on most current systems, but they render your code non-portable. (Note that it's common on 64-bit systems for void* and int to be different sizes, so printing pointers with %d" is really non-portable, not just theoretically.)
Incidentally, the output format for "%p" is implementation-defined. Hexadecimal is common, (in upper or lower case, with or without a leading "0x" or "0X"), but it's not the only possibility. All you can count on is that, assuming a reasonable implementation, it will be a reasonable way to represent a pointer value in human-readable form (and that scanf will understand the output of printf).
The article you read is entirely correct. The correct way to print an int* value is
printf("%p", (void*)&a);
Don't take the lazy way out; it's not at all difficult to get it right.
Suggested reading: Section 4 of the comp.lang.c FAQ. (Further suggested reading: All the other sections.
EDIT:
In response to Alcott's question:
There is still one thing I don't quite understand. int a = 10; int *p = &a;, so p's value is a's address in mem, right? If right, then p's value will range from 0 to 2^32-1 (if cpu is 32-bit), and an integer is 4-byte on 32-bit OS, right? then What's the difference between the p's value and an integer? Can p's value go out of the range?
The difference is that they're of different types.
Assume a system on which int, int*, void*, and float are all 32 bits (this is typical for current 32-bit systems). Does the fact that float is 32 bits imply that its range is 0 to 232-1? Or -231 to 231-1? Certainly not; the range of float (assuming IEEE representation) is approximately -3.40282e+38 to +3.40282e+38, with widely varying resolution across the range, plus exotic values like negative zero, subnormalized numbers, denormalized numbers, infinities, and NaNs (Not-a-Number). int and float are both 32 bits, and you can take the 32 bits of a float object and treat it as an int representation, but the result won't have any straightforward relationship to the value of the float. The second low-order bit of an int, for example, has a specific meaning; it contributes 0 to the value if it's 0, and 2 to the value if it's 1; the corresponding bit of a float has a meaning, but it's quite different (it contributes a value that depends on the value of the exponent).
The situation with pointers is quite similar. A pointer value has a meaning: it's the address of some object (or any of several other things, but we'll set that aside for now). On most current systems, interpreting the bits of a pointer object as if it were an integer gives you something that makes sense on the machine level. But the language itself does not guarantee, or even hint, that that's the case.
Pointers are not numbers.
A concrete example: some years ago, I ran across some code that tried to compute the difference in bytes between two addresses by casting to integers. It was something like this:
unsigned char *p0;
unsigned char *p1;
long difference = (unsigned long)p1 - (unsigned long)p0;
If you assume that pointers are just numbers, representing addresses in a linear monolithic address space, then this code makes sense. But that assumption is not supported by the language. And in fact, there was a system on which that code was intended to run (the Cray T90) on which it simply would not have worked. The T90 had 64-bit pointers pointing to 64-bit words. Byte pointers were synthesized in software by storing an offset in the 3 high-order bits of a pointer object. Subtracting two pointers in the above manner, if they both had 0 offsets, would give you the number of words, not bytes, between the addresses. And if they had non-0 offsets, it would give you meaningless garbage. (Conversion from a pointer to an integer would just copy the bits; it could have done the work to give you a meaningful byte index, but it didn't.)
The solution was simple: drop the casts and use pointer arithmetic:
long difference = p1 - p0;
Other addressing schemes are possible. For example, an address might consist of a descriptor that (perhaps indirectly) references a block of memory, plus an offset within that block.
You can assume that addresses are just numbers, that the address space is linear and monolithic, that all pointers are the same size and have the same representation, that a pointer can be safely converted to int, or to long, and back again without loss of information. And the code you write based on those assumptions will probably work on most current systems. But it's entirely possible that some future systems will again use a different memory model, and your code will break.
If you avoid making any assumptions beyond what the language actually guarantees, your code will be far more future-proof. And even leaving portability issues aside, it will probably be cleaner.
So much insanity present here...
%p is generally the correct format specifier to use if you just want to print out a representation of the pointer. Never, ever use %d.
The length of an int and the length of a pointer (void* or otherwise) have no relationship. Most data models on i386 just happen to have 32-bit ints AND 32-bit pointers -- other platforms, including x86-64, are not the same! (This is also historically known as "all the world's a VAX syndrome".) http://en.wikipedia.org/wiki/64-bit#64-bit_data_models
If for some reason you want to hold a memory address in an integral variable, use the right types! intptr_t and uintptr_t. They're in stdint.h. See http://en.wikipedia.org/wiki/Stdint.h#Integers_wide_enough_to_hold_pointers
In C void * is an un-typed pointer. void does not mean void... it means anything. Thus casting to void * would be the same as casting to "pointer" in another language.
Using (int *)&a should work too... but the stylistic point of saying (void *) is to say -- I don't care about the type -- just that it is a pointer.
Note: It is possible for an implementation of C to cause this construct to fail and still meet the requirements of the standards. I don't know of any such implementations, but it is possible.
Although it the vast majority of C implementations store pointers to all kinds of objects using the same representation, the C Standard does not require that all implementations do so, nor does it even provide any means by which a program which would exploit commonality of representations could test whether an implementation follows the common practice and refuse to run if an implementation doesn't.
If on some particular platform, an int* held a word address, while both char* and void* combine a word address with a word that identifies a byte within a word, passing an int* to a function that is expecting to retrieve a variadic argument of type char* or void* would result in that function trying to fetch more data from the stack (a word address plus the supplemental word) than had been pushed (just the word address). This could cause the system to malfunction in unpredictable ways.
Many compilers for commonplace platforms that use the same representation for all pointers will process an action which passes a non-void pointer precisely the same way as they would process an action which casts the pointer to void* before passing it. They thus have no reason to care about whether the pointer type that is passed as a variadic argument will precisely match the pointer type expected by the recipient. Although the Standard could have specified that such implementations which would have no reason to care about pointer types should behave as though the pointers were cast to void*, the authors of C89 Standard avoided describing anything which wouldn't be common to all conforming compilers. The Standard's terminology for a construct that 99% of implementations should process identically, but 1% would might process unpredictably, is "Undefined Behavior". Implementations may, and often should, extend the semantics of the language by specifying how they will treat such constructs, but that's a Quality of Implementation issue outside the Standard's jurisdiction.
Related
I've been doing some pointers testing in C, and I was just curious if the addresses of a function's parameters are always in a difference of 4 bytes from one another.
I've tries to run the following code:
#include <stdio.h>
void func(long a, long b);
int main(void)
{
func(1, 2);
getchar();
return 0;
}
void func(long a, long b)
{
printf("%d\n", (int)&b - (int)&a);
}
This code seems to always print 4, no matter what is the type of func's parameters.
I was just wondering if it's ALWAYS 4, because if so it can be useful for something I'm trying to do (but if it isn't necessarily 4 I guess I could just use va_list for my function or something).
So: Is it necessarily 4 bytes?
Absolutely not, in so many ways that it would be hard to count them all.
First and foremost, the memory layout of arguments is simply not specified by the C language. Full stop. It is not specified. Thus the answer is "no" immediately.
va_list exists because there was a need to be able to navigate a list of varadic arguments because it wasn't specified other than that. va_list is intentionally very limited, so that it works on platforms where the shape of the stack does not match your intuition.
Other reasons it can't always be 4:
What if you pass an object of length 8?
What if the compiler optimizes a reference to actually point at the object in another frame?
What if the compiler adds padding, perhaps to align a 64-bit number on a 64-bit boundary?
What if the stack is built in the opposite direction (such that the difference would be -4 instead of +4)
The list goes on and on. C does not specify the relative addresses between arguments.
As the other answers correctly say:
No.
Furthermore, even trying to determine whether the addresses differ by 4 bytes, depending on how you do it, probably has undefined behavior, which means the C standard says nothing about what your program does.
void func(long a, long b)
{
printf("%d\n", (int)&b - (int)&a);
}
&a and &b are expression of type long*. Converting a pointer to int is legal, but the result is implementation-defined, and "If the result cannot be represented in the integer type, the behavior is undefined. The result need not be in the range of values of any integer type."
It's very likely that pointers are 64 bits and int is 32 bits, so the conversions could lose information.
Most likely the conversions will give you values of type int, but they don't necessarily have any meaning, nor does their difference.
Now you can subtract pointer values directly, with a result of the signed integer type ptrdiff_t (which, unlike int, is probably big enough to hold the result).
printf("%td\n", &b - &a);
But "When two pointers are subtracted, both shall point to elements of the same array object, or one past the last element of the array object; the result is the difference of the subscripts of the two array elements." Pointers to distinct object cannot be meaningfully compared or subtracted.
Having said all that, it's likely that the implementation you're using has a memory model that's reasonably straightforward, and that pointer values are in effect represented as indices into a monolithic memory space. Comparing &b vs. &a is not permitted by the C language, but examining the values can provide some insight about what's going on behind the curtain -- which can be especially useful if you're tracking down a bug.
Here's something you can do portably to examine the addresses:
printf("&a = %p\n", (void*)&a);
printf("&b = %p\n", (void*)&b);
The result you're seeing for the subtraction (4) suggests that type long is probably 4 bytes (32 bits) on your system. I'd guess you're on Windows. It also suggests something about the way function parameters are allocated -- something that you as a programmer should almost never have to care about, but is worth understanding anyway.
[...] I was just curious if the addresses of a function's parameters are always in a difference of 4 bytes from one another."
The greatest error in your reasoning is to think that the parameters exist in memory at all.
I am running this program on x86-64:
#include <stdio.h>
#include <stdint.h>
void func(long a, long b)
{
printf("%d\n", (int)((intptr_t)&b - (intptr_t)&a));
}
int main(void)
{
func(1, 2);
}
and compile it with gcc -O3 it prints 8, proving that your guess is absolutely wrong. Except... when I compile it without optimization it prints out -8.
X86-64 SYSV calling convention says that the arguments are passed in registers instead of being passed in memory. a and b do not have an address until you take their address with & - then the compiler is caught with its pants down from cheating the as-if rule and it quickly pulls up its pants and stuffs them into some memory location so that they can have their address taken, but it is in no way consistent in where they're stored.
I am creating an emulator for an instruction set architecture, and I needed to implement a stack structure. I decided that my %eip, %ebp and %esp would be int pointers. However, there are situations where I need to store memory addresses on the stack, in which case this memory would be encoded as an integer value. But when I return this value, I need to put it back into my instruction pointer, which is implemented as an int pointer. C will not let me assign my integer to my int pointer, so I have no way of recovering these memory addresses from the "stack". Any suggestions?
To assign an int value to an int * object, use an explicit cast, as in:
destination = (int *) source;
Your question says “C will not let me assign my integer to my int pointer” but fails to state exactly what the problem is. Presumably you are getting some diagnostic message from the compiler. This would be because assigning an int value to an int * object violates the C standard’s constraints for assignments. The code above shows how to work around that.
That solves the immediate problem of the compiler diagnostic. However, there can be various issues with using int values as containers for pointers, including the possibility of trap values and discrepancies between the sizes of pointers and integers. Provided that int and int * are the same size, using an int to hold an int * is not unlikely to work, but you should be sure of the properties of your C implementation.
I decided that my %eip, %ebp and %esp would be int pointers.
This is not a sound architectural decision. You need to reconsider it.
The size of a pointer is architecture-dependent -- in particular, an int * will be 64 bits wide on a 64-bit system. By contrast, all of these registers are 32 bits wide by definition. Using a 64-bit pointer to store their values will result in unexpected behavior.
These registers are not required to be aligned to an integer. In particular, EIP is (at best) aligned to an instruction, and will be incremented by one byte when running 1-byte instructions. Deferencing an int * which is not properly aligned will cause an unaligned access fault on many systems.
There is no hard architectural distinction between any of the integer registers (EA/B/C/DX, ESP, EBP, ESI, EDI). All of them can be referenced in an ModRM encoding, and can be treated as either a numeric value or an address, depending on the context. Singling ESP and EBP out will unnecessarily complicate your emulator, and is likely to create a lot of obnoxious special cases in your code.
Note that, as you are emulating a 32-bit system on what might not be a 32-bit platform, you will need some way of translating addresses within the emulated system to "real" addresses in the host process. There are a number of different ways of doing this; which one is most appropriate for you will depend on your specific goals.
It is implementation defined but if the integer width is not smaller than the pointer - you can use it this way.
Some people say that the using ptrdiff_t and NULL pointer as a reference is more portable and safer.
ptrdiff_t myptrdiff = myptr - (type_of_myptr *)NULL;
myptr = myptrdiff + (type_of_myptr *)NULL;
Hi I'm sure this must be a common question but I can't find the answer when I search for it. My question basically concerns two pointers. I want to compare their addresses and determine if one is bigger than the other. I would expect all addresses to be unsigned during comparison. Is this true, and does it vary between C89, C99 and C++? When I compile with gcc the comparison is unsigned.
If I have two pointers that I'm comparing like this:
char *a = (char *) 0x80000000; //-2147483648 or 2147483648 ?
char *b = (char *) 0x1;
Then a is greater. Is this guaranteed by a standard?
Edit to update on what I am trying to do. I have a situation where I would like to determine that if there's an arithmetic error it will not cause a pointer to go out of bounds. Right now I have the start address of the array and the end address. And if there's an error and the pointer calculation is wrong, and outside of the valid addresses of memory for the array, I would like to make sure no access violation occurs. I believe I can prevent this by comparing the suspect pointer, which has been returned by another function, and determining if it is within the acceptable range of the array. The question of negative and positive addresses has to do with whether I can make the comparisons, as discussed above in my original question.
I appreciate the answers so far. Based on my edit would you say that what I'm doing is undefined behavior in gcc and msvc? This is a program that will run on Microsoft Windows only.
Here's an over simplified example:
char letters[26];
char *do_not_read = &letters[26];
char *suspect = somefunction_i_dont_control(letters,26);
if( (suspect >= letters) && (suspect < do_not_read) )
printf("%c", suspect);
Another edit, after reading AndreyT's answer it appears to be correct. Therefore I will do something like this:
char letters[26];
uintptr_t begin = letters;
uintptr_t toofar = begin + sizeof(letters);
char *suspect = somefunction_i_dont_control(letters,26);
if( ((uintptr_t)suspect >= begin) && ((uintptr_t)suspect < toofar ) )
printf("%c", suspect);
Thanks everyone!
Pointer comparisons cannot be signed or unsigned. Pointers are not integers.
C language (as well as C++) defines relative pointer comparisons only for pointers that point into the same aggregate (struct or array). The ordering is natural: the pointer that points to an element with smaller index in an array is smaller. The pointer that points to a struct member declared earlier is smaller. That's it.
You can't legally compare arbitrary pointers in C/C++. The result of such comparison is not defined. If you are interested in comparing the numerical values of the addresses stored in the pointers, it is your responsibility to manually convert the pointers to integer values first. In that case, you will have to decide whether to use a signed or unsigned integer type (intptr_t or uintptr_t). Depending on which type you choose, the comparison will be "signed" or "unsigned".
The integer-to-pointer conversion is wholly implementation defined, so it depends on the implementation you are using.
That said, you are only allowed to relationally compare pointers that point to parts of the same object (basically, to subobjects of the same struct or elements of the same array). You aren't allowed to compare two pointers to arbitrary, wholly unrelated objects.
From a draft C++ Standard 5.9:
If two pointers p and q of the same type point to different objects
that are not members of the same object or elements of the same array
or to different functions, or if only one of them is null, the results
of p<q, p>q, p<=q, and p>=q are unspecified.
So, if you cast numbers to pointers and compare them, C++ gives you unspecified results. If you take the address of elements you can validly compare, the results of comparison operations are specified independently of the signed-ness of the pointer types.
Note unspecified is not undefined: it's quite possible to compare pointers to different objects of the same type that aren't in the same structure or array, and you can expect some self-consistent result (otherwise it'd be impossible to use such pointers as keys in trees, or to sort a vector of such pointers, binary search the vector etc., where a consistent intuitive overall < ordering is needed).
Note that in very old C++ Standards the behaviour was undefined - like the 2005 WG14/N1124 draft andrewdski links to under James McNellis's answer -
To complement the other answers, comparison between pointers that point to different objects depends on the standard.
In C99 (ISO/IEC 9899:1999 (E)), §6.5.8:
5 [...] In all other cases, the behavior is undefined.
In C++03 (ISO/IEC 14882:2003(E)), §5.9:
-Other pointer comparisons are unspecified.
I know several of the answers here say you cannot compare pointers unless they point to within the same structure, but that's a red herring and I'll try to explain why. One of your pointers points to the start of your array, the other to the end, so they are pointing to the same structure. A language lawyer could say that if your third pointer points outside of the object, the comparison is undefined, so x >= array.start might be true for all x. But this is no issue, since at the point of comparison C++ cannot know if the array isn't embedded in an even bigger structure. Furthermore, if your address space is linear, like it's bound to be these days, your pointer comparison will be implemented as an (un)signed integer comparison, since any other implementation would be slower. Even in the times of segments and offsets, (far) pointer comparison was implemented by first normalising the pointer and then comparing them as integers.
What this all boils down to then, is that if your compiler is okay, comparing the pointers without worrying about the signs should work, if all you care about is that the pointer points within the array, since the compiler should make the pointers signed or unsigned depending on which of the two boundaries a C++ object may straddle.
Different platforms behave differently in this matter, which is why C++ has to leave it up to the platform. There are even platforms in which both addresses near 0 and 80..00h are not mappable or already taken at process start-up. In that case, it doesn't matter, as long as you're consistent about it.
Sometimes this can cause compatibility issues. As an example, in Win32 pointers are unsigned. Now, it used to be the case that of the 4GB address space only the lower half (more precisely 10000h ... 7FFFFFFFh, because of the NULL-Pointer Assignment Partition) was available to applications; high addresses were only available to the kernel. This caused some people to put addresses in signed variables, and their programs would keep working since the high bit was always 0. But then came /3GB switch, which made almost 3 GB available to applications (more precisely 10000h ... BFFFFFFFh) and the application would crash or behave erratically.
You explicitly state your program will be Windows-only, which uses unsigned pointers. However, maybe you'll change your mind in the future, and using intptr_t or uintptr_t is bad for portability. I also wonder if you should be doing this at all... if you're indexing into an array it might be safer to compare indices instead. Suppose for example that you have a 1 GB array at 1500000h ... 41500000h, consisting of 16,384 elements of 64 kB each. Suppose you accidentally look up index 80,000 – clearly out of range. The pointer calculation will yield 39D00000h, so your pointer check will allow it, even though it shouldn't.
I have a function that I would like to be able to return special values for failure and uninitialized (it returns a pointer on success).
Currently it returns NULL for failure, and -1 for uninitialized, and this seems to work... but I could be cheating the system. IIRC, addresses are always positive, are they not? (although since the compiler is allowing me to set an address to -1, this seems strange).
[update]
Another idea I had (in the event that -1 was risky) is to malloc a char # the global scope, and use that address as a sentinel.
No, addresses aren't always positive - on x86_64, pointers are sign-extended and the address space is clustered symmetrically around 0 (though it is usual for the "negative" addresses to be kernel addresses).
However the point is mostly moot, since C only defines the meaning of < and > pointer comparisons between pointers that are to part of the same object, or one past the end of an array. Pointers to completely different objects cannot be meaningfully compared other than for exact equality, at least in standard C - if (p < NULL) has no well defined semantics.
You should create a dummy object with static storage duration and use its address as your unintialised value:
extern char uninit_sentinel;
#define UNINITIALISED ((void *)&uninit_sentinel)
It's guaranteed to have a single, unique address across your program.
The valid values for a pointer are entirely implementation-dependent, so, yes, a pointer address could be negative.
More importantly, however, consider (as an example of a possible implementation choice) the case where you are on a 32-bit platform with a 32-bit pointer size. Any value that can be represented by that 32-bit value might be a valid pointer. Other than the null pointer, any pointer value might be a valid pointer to an object.
For your specific use case, you should consider returning a status code and perhaps taking the pointer as a parameter to the function.
It's generally a bad design to try to multiplex special values onto a return value... you're trying to do too much with a single value. It would be cleaner to return your "success pointer" via argument, rather than the return value. That leaves lots of non-conflicting space in the return value for all of the conditions you want to describe:
int SomeFunction(SomeType **p)
{
*p = NULL;
if (/* check for uninitialized ... */)
return UNINITIALIZED;
if (/* check for failure ... */)
return FAILURE;
*p = yourValue;
return SUCCESS;
}
You should also do typical argument checking (ensure that 'p' isn't NULL).
The C language does not define the notion of "negativity" for pointers. The property of "being negative" is a chiefly arithmetical one, not in any way applicable to values of pointer type.
If you have a pointer-returning function, then you cannot meaningfully return the value of -1 from that function. In C language integral values (other than zero) are not implicitly convertible to pointer types. An attempt to return -1 from a pointer-returning function is an immediate constraint violation that will result in diagnostic message. In short, it is an error. If your compiler allows it, it simply means that it doesn't enforce that constraint too strictly (most of the time they do it for compatibility with pre-standard code).
If you force the value of -1 to pointer type by an explicit cast, the result of the cast will be implementation-defined. The language itself makes no guarantees about it. It might easily prove to be the same as some other, valid pointer value.
If you want to create a reserved pointer value, there no need to malloc anything. You can simple declare a global variable of the desired type and use its address as the reserved value. It is guaranteed to be unique.
Pointers can be negative like an unsigned integer can be negative. That is, sure, in a two's-complement interpretation, you could interpret the numerical value to be negative because the most-significant-bit is on.
What's the difference between failure and unitialized. If unitialized is not another kind of failure, then you probably want to redesign the interface to separate these two conditions.
Probably the best way to do this is to return the result through a parameter, so the return value only indicates an error. For example where you would write:
void* func();
void* result=func();
if (result==0)
/* handle error */
else if (result==-1)
/* unitialized */
else
/* initialized */
Change this to
// sets the *a to the returned object
// *a will be null if the object has not been initialized
// returns true on success, false otherwise
int func(void** a);
void* result;
if (func(&result)){
/* handle error */
return;
}
/*do real stuff now*/
if (!result){
/* initialize */
}
/* continue using the result now that it's been initialized */
#James is correct, of course, but I'd like to add that pointers don't always represent absolute memory addresses, which theoretically would always be positive. Pointers also represent relative addresses to some point in memory, often a stack or frame pointer, and those can be both positive and negative.
So your best bet is to have your function accept a pointer to a pointer as a parameter and fill that pointer with a valid pointer value on success while returning a result code from the actual function.
James answer is probably correct, but of course describes an implementation choice, not a choice that you can make.
Personally, I think addresses are "intuitively" unsigned. Finding a pointer that compares as less-than a null pointer would seem wrong. But ~0 and -1, for the same integer type, give the same value. If it's intuitively unsigned, ~0 may make a more intuitive special-case value - I use it for error-case unsigned ints quite a lot. It's not really different (zero is an int by default, so ~0 is -1 until you cast it) but it looks different.
Pointers on 32-bit systems can use all 32 bits BTW, though -1 or ~0 is an extremely unlikely pointer to occur for a genuine allocation in practice. There are also platform-specific rules - for example on 32-bit Windows, a process can only have a 2GB address space, and there's a lot of code around that encodes some kind of flag into the top bit of a pointer (e.g. for balancing flags in balanced binary trees).
Actually, (at least on x86), the NULL-pointer exception is generated not only by dereferencing the NULL pointer, but by a larger range of addresses (eg, first 65kb). This helps catching such errors as
int* x = NULL;
x[10] = 1;
So, there are more addresses that are garanteed to generate the NULL pointer exception when dereferenced.
Now consider this code (made compilable for AndreyT):
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define ERR_NOT_ENOUGH_MEM (int)NULL
#define ERR_NEGATIVE (int)NULL + 1
#define ERR_NOT_DIGIT (int)NULL + 2
char* fn(int i){
if (i < 0)
return (char*)ERR_NEGATIVE;
if (i >= 10)
return (char*)ERR_NOT_DIGIT;
char* rez = (char*)malloc(strlen("Hello World ")+sizeof(char)*2);
if (rez)
sprintf(rez, "Hello World %d", i);
return rez;
};
int main(){
char* rez = fn(3);
switch((int)rez){
case ERR_NOT_ENOUGH_MEM: printf("Not enough memory!\n"); break;
case ERR_NEGATIVE: printf("The parameter was negative\n"); break;
case ERR_NOT_DIGIT: printf("The parameter is not a digit\n"); break;
default: printf("we received %s\n", rez);
};
return 0;
};
this could be useful in some cases.
It won't work on some Harvard architectures, but will work on von Neumann ones.
Do not use malloc for this purpose. It might keep unnecessary memory tied up (if a lot of memory is already in use when malloc gets called and the sentinel gets allocated at a high address, for example) and it confuses memory debuggers/leak detectors. Instead simply return a pointer to a local static const char object. This pointer will never compare equal to any pointer the program could obtain in any other way, and it only wastes one byte of bss.
You don't need to care about the signness of a pointer, because it's implementation defined. The real question here is "how to return special values from a function returning pointer?" which I've explained in detail in my answer to the question Pointer address span on various platforms
In summary, the all-one bit pattern (-1) is (almost) always safe, because it's already at the end of the spectrum and data cannot be stored wrapped around to the first address, and the malloc family never returns -1. In fact this value is even returned by many Linux system calls and Win32 APIs to indicate another state for the pointer. So if you need just failure and uninitialized then it's a good choice
But you can return far more error states by utilizing the fact that variables must be aligned properly (unless you specified some other options). For example in a pointer to int32_t the low 2 bits are always zero which means only ¹⁄₄ of the possible values are valid addresses, leaving all of the remaining bit patterns for you to use. So a simple solution would be just checking the lowest bit
int* result = func();
if (!result)
error_happened();
else if ((uintptr_t)result & 1)
uninitialized();
In this case you can return both a valid pointer and some additional data at the same time
You can also use the high bits for storing data in 64-bit systems. On ARM there's a flag that tells the CPU to ignore the high bits in the addresses. On x86 there isn't a similar thing but you can still use those bits as long as you make it canonical before dereferencing. See Using the extra 16 bits in 64-bit pointers
See also
Is ((void *) -1) a valid address?
NULL is the only valid error return in this case, this is true anytime an unsigned value such as a pointer is returned. It may be true that in some cases pointers will not be large enough to use the sign bit as a data bit, however since pointers are controlled by the OS not the program I would not rely on this behavior.
Remember that a pointer is basically a 32-bit value; whether or not this is a possible negative or always positive number is just a matter of interpretation (i.e.) whether the 32nd bit is interpreted as the sign bit or as a data bit. So if you interpreted 0xFFFFFFF as a signed number it would be -1, if you interpreted it as an unsigned number it would be 4294967295. Technically, it is unlikely that a pointer would ever be this large, but this case should be considered anyway.
As far as an alternative you could use an additional out parameter (returning NULL for all failures), however this would require clients to create and pass a value even if they don't need to distinguish between specific errors.
Another alternative would be to use the GetLastError/SetLastError mechanism to provide additional error information (This would be specific to Windows, don't know if that is an issue or not), or to throw an exception on error instead.
Positive or negative is not a meaningful facet of pointer type. They pertain to signed integer including signed char, short, int etc.
People talk about negative pointer mostly in a situation that treats pointer's machine representation as an integer type. e.g. reinterpret_cast<intptr_t>(ptr). In this case, they are actually talking about the cast integer, not the pointer itself.
In some scenario I think pointer is inherently unsigned, we talk about address in terms below or above. 0xFFFF.FFFF is above 0x0AAAA.0000, which is intuitively for human beings. Although 0xFFFF.FFFF is actually a "negative" while 0x0AAA.0000 is positive.
But in other scenarios such as pointer subtraction (ptr1 - ptr2) that results in a signed value whose type is ptrdiff_t, it's inconsistent when you compare with integer's subtraction, signed_int_a - signed_int_b results in a signed int type, unsigned_int_a - unsigned_int_b produces an unsigned type. But for pointer subtraction, it produces a signed type, because the semantic is the distance between two pointers, the unit is number of elements.
In summary I suggest treating pointer type as standalone type, every type has it's set of operation on it. For pointers (excluding function pointer, member function pointer, and void *):
List item
+, +=
ptr + any_integer_type
-, -=
ptr - any_integer_type
ptr1 - ptr2
++ both prefix and postfix
-- both prefix and postfix
Note there are no / * % operations for pointer. That's also supported that pointer should be treated as a standalone type, instead of "A type similar to int" or "A type whose underlying type is int so it should looks like int".
I would like to know architectures which violate the assumptions I've listed below. Also, I would like to know if any of the assumptions are false for all architectures (that is, if any of them are just completely wrong).
sizeof(int *) == sizeof(char *) == sizeof(void *) == sizeof(func_ptr *)
The in-memory representation of all pointers for a given architecture is the same regardless of the data type pointed to.
The in-memory representation of a pointer is the same as an integer of the same bit length as the architecture.
Multiplication and division of pointer data types are only forbidden by the compiler. NOTE: Yes, I know this is nonsensical. What I mean is - is there hardware support to forbid this incorrect usage?
All pointer values can be casted to a single integer. In other words, what architectures still make use of segments and offsets?
Incrementing a pointer is equivalent to adding sizeof(the pointed data type) to the memory address stored by the pointer. If p is an int32* then p+1 is equal to the memory address 4 bytes after p.
I'm most used to pointers being used in a contiguous, virtual memory space. For that usage, I can generally get by thinking of them as addresses on a number line. See Stack Overflow question Pointer comparison.
I can't give you concrete examples of all of these, but I'll do my best.
sizeof(int *) == sizeof(char *) == sizeof(void *) == sizeof(func_ptr *)
I don't know of any systems where I know this to be false, but consider:
Mobile devices often have some amount of read-only memory in which program code and such is stored. Read-only values (const variables) may conceivably be stored in read-only memory. And since the ROM address space may be smaller than the normal RAM address space, the pointer size may be different as well. Likewise, pointers to functions may have a different size, as they may point to this read-only memory into which the program is loaded, and which can otherwise not be modified (so your data can't be stored in it).
So I don't know of any platforms on which I've observed that the above doesn't hold, but I can imagine systems where it might be the case.
The in-memory representation of all pointers for a given architecture is the same regardless of the data type pointed to.
Think of member pointers vs regular pointers. They don't have the same representation (or size). A member pointer consists of a this pointer and an offset.
And as above, it is conceivable that some CPU's would load constant data into a separate area of memory, which used a separate pointer format.
The in-memory representation of a pointer is the same as an integer of the same bit length as the architecture.
Depends on how that bit length is defined. :)
An int on many 64-bit platforms is still 32 bits. But a pointer is 64 bits.
As already said, CPU's with a segmented memory model will have pointers consisting of a pair of numbers. Likewise, member pointers consist of a pair of numbers.
Multiplication and division of pointer data types are only forbidden by the compiler.
Ultimately, pointers data types only exist in the compiler. What the CPU works with is not pointers, but integers and memory addresses. So there is nowhere else where these operations on pointer types could be forbidden. You might as well ask for the CPU to forbid concatenation of C++ string objects. It can't do that because the C++ string type only exists in the C++ language, not in the generated machine code.
However, to answer what you mean, look up the Motorola 68000 CPUs. I believe they have separate registers for integers and memory addresses. Which means that they can easily forbid such nonsensical operations.
All pointer values can be casted to a single integer.
You're safe there. The C and C++ standards guarantee that this is always possible, no matter the memory space layout, CPU architecture and anything else. Specifically, they guarantee an implementation-defined mapping. In other words, you can always convert a pointer to an integer, and then convert that integer back to get the original pointer. But the C/C++ languages say nothing about what the intermediate integer value should be. That is up to the individual compiler, and the hardware it targets.
Incrementing a pointer is equivalent to adding sizeof(the pointed data type) to the memory address stored by the pointer.
Again, this is guaranteed. If you consider that conceptually, a pointer does not point to an address, it points to an object, then this makes perfect sense. Adding one to the pointer will then obviously make it point to the next object. If an object is 20 bytes long, then incrementing the pointer will move it 20 bytes, so that it moves to the next object.
If a pointer was merely a memory address in a linear address space, if it was basically an integer, then incrementing it would add 1 to the address -- that is, it would move to the next byte.
Finally, as I mentioned in a comment to your question, keep in mind that C++ is just a language. It doesn't care which architecture it is compiled to. Many of these limitations may seem obscure on modern CPU's. But what if you're targeting yesteryear's CPU's? What if you're targeting the next decade's CPU's? You don't even know how they'll work, so you can't assume much about them. What if you're targeting a virtual machine? Compilers already exist which generate bytecode for Flash, ready to run from a website. What if you want to compile your C++ to Python source code?
Staying within the rules specified in the standard guarantees that your code will work in all these cases.
I don't have specific real world examples in mind but the "authority" is the C standard. If something is not required by the standard, you can build a conforming implementation that intentionally fails to comply with any other assumptions. Some of these assumption are true most of the time just because it's convenient to implement a pointer as an integer representing a memory address that can be directly fetched by the processor but this is just a consequent of "convenience" and can't be held as a universal truth.
Not required by the standard (see this question). For instance, sizeof(int*) can be unequal to size(double*). void* is guaranteed to be able to store any pointer value.
Not required by the standard. By definition, size is a part of representation. If the size can be different, the representation can be different too.
Not necessarily. In fact, "the bit length of an architecture" is a vague statement. What is a 64-bit processor, really? Is it the address bus? Size of registers? Data bus? What?
It doesn't make sense to "multiply" or "divide" a pointer. It's forbidden by the compiler but you can of course multiply or divide the underlying representation (which doesn't really make sense to me) and that results in undefined behavior.
Maybe I don't understand your point but everything in a digital computer is just some kind of binary number.
Yes; kind of. It's guaranteed to point to a location that's a sizeof(pointer_type) farther. It's not necessarily equivalent to arithmetic addition of a number (i.e. farther is a logical concept here. The actual representation is architecture specific)
For 6.: a pointer is not necessarily a memory address. See for example "The Great Pointer Conspiracy" by Stack Overflow user jalf:
Yes, I used the word “address” in the comment above. It is important to realize what I mean by this. I do not mean “the memory address at which the data is physically stored”, but simply an abstract “whatever we need in order to locate the value. The address of i might be anything, but once we have it, we can always find and modify i."
And:
A pointer is not a memory address! I mentioned this above, but let’s say it again. Pointers are typically implemented by the compiler simply as memory addresses, yes, but they don’t have to be."
Some further information about pointers from the C99 standard:
6.2.5 §27 guarantees that void* and char* have identical representations, ie they can be used interchangably without conversion, ie the same address is denoted by the same bit pattern (which doesn't have to be true for other pointer types)
6.3.2.3 §1 states that any pointer to an incomplete or object type can be cast to (and from) void* and back again and still be valid; this doesn't include function pointers!
6.3.2.3 §6 states that void* can be cast to (and from) integers and 7.18.1.4 §1 provides apropriate types intptr_t and uintptr_t; the problem: these types are optional - the standard explicitly mentions that there need not be an integer type large enough to actually hold the value of the pointer!
sizeof(char*) != sizeof(void(*)(void) ? - Not on x86 in 36 bit addressing mode (supported on pretty much every Intel CPU since Pentium 1)
"The in-memory representation of a pointer is the same as an integer of the same bit length" - there's no in-memory representation on any modern architecture; tagged memory has never caught on and was already obsolete before C was standardized. Memory in fact doesn't even hold integers, just bits and arguably words (not bytes; most physical memory doesn't allow you to read just 8 bits.)
"Multiplication of pointers is impossible" - 68000 family; address registers (the ones holding pointers) didn't support that IIRC.
"All pointers can be cast to integers" - Not on PICs.
"Incrementing a T* is equivalent to adding sizeof(T) to the memory address" - true by definition. Also equivalent to &pointer[1].
I don't know about the others, but for DOS, the assumption in #3 is untrue. DOS is 16 bit and uses various tricks to map many more than 16 bits worth of memory.
The in-memory representation of a pointer is the same as an integer of the same bit length as the architecture.
I think this assumption is false because on the 80186, for example, a 32-bit pointer is held in two registers (an offset register an a segment register), and which half-word went in which register matters during access.
Multiplication and division of pointer data types are only forbidden by the compiler.
You can't multiply or divide types. ;P
I'm unsure why you would want to multiply or divide a pointer.
All pointer values can be casted to a single integer. In other words, what architectures still make use of segments and offsets?
The C99 standard allows pointers to be stored in intptr_t, which is an integer type. So, yes.
Incrementing a pointer is equivalent to adding sizeof(the pointed data type) to the memory address stored by the pointer. If p is an int32* then p+1 is equal to the memory address 4 bytes after p.
x + y where x is a T * and y is an integer is equivilent to (T *)((intptr_t)x + y * sizeof(T)) as far as I know. Alignment may be an issue, but padding may be provided in the sizeof. I'm not really sure.
In general, the answer to all of the questions is "yes", and it's because only those machines that implement popular languages directly saw the light of day and persisted into the current century. Although the language standards reserve the right to vary these "invariants", or assertions, it hasn't ever happened in real products, with the possible exception of items 3 and 4 which require some restatement to be universally true.
It's certainly possible to build segmented MMU designs, which correspond roughly with the capability-based architectures that were popular academically in past years, but no such system has typically seen common use with such features enabled. Such a system might have conflicted with the assertions as it would probably have had large pointers.
In addition to segmented/capability MMUs, which often have large pointers, more extreme designs have tried to encode data types in pointers. Few of these were ever built. (This question brings up all of the alternatives to the basic word-oriented, a pointer-is-a-word architectures.)
Specifically:
The in-memory representation of all pointers for a given architecture is the same regardless of the data type pointed to. True except for extremely wacky past designs that tried to implement protection not in strongly-typed languages but in hardware.
The in-memory representation of a pointer is the same as an integer of the same bit length as the architecture. Maybe, certainly some sort of integral type is the same, see LP64 vs LLP64.
Multiplication and division of pointer data types are only forbidden by the compiler. Right.
All pointer values can be casted to a single integer. In other words, what architectures still make use of segments and offsets? Nothing uses segments and offsets today, but a C int is often not big enough, you may need a long or long long to hold a pointer.
Incrementing a pointer is equivalent to adding sizeof(the pointed data type) to the memory address stored by the pointer. If p is an int32* then p+1 is equal to the memory address 4 bytes after p. Yes.
It is interesting to note that every Intel Architecture CPU, i.e., every single PeeCee, contains an elaborate segmentation unit of epic, legendary, complexity. However, it is effectively disabled. Whenever a PC OS boots up, it sets the segment bases to 0 and the segment lengths to ~0, nulling out the segments and giving a flat memory model.
There were lots of "word addressed" architectures in the 1950s, 1960s and 1970s. But I cannot recall any mainstream examples that had a C compiler. I recall the ICL / Three Rivers PERQ machines in the 1980s that was word addressed and had a writable control store (microcode). One of its instantiations had a C compiler and a flavor of Unix called PNX, but the C compiler required special microcode.
The basic problem is that char* types on word addressed machines are awkward, however you implement them. You often up with sizeof(int *) != sizeof(char *) ...
Interestingly, before C there was a language called BCPL in which the basic pointer type was a word address; that is, incrementing a pointer gave you the address of the next word, and ptr!1 gave you the word at ptr + 1. There was a different operator for addressing a byte: ptr%42 if I recall.
EDIT: Don't answer questions when your blood sugar is low. Your brain (certainly, mine) doesn't work as you expect. :-(
Minor nitpick:
p is an int32* then p+1
is wrong, it needs to be unsigned int32, otherwise it will wrap at 2GB.
Interesting oddity - I got this from the author of the C compiler for the Transputer chip - he told me that for that compiler, NULL was defined as -2GB. Why? Because the Transputer had a signed address range: -2GB to +2GB. Can you beleive that? Amazing isn't it?
I've since met various people that have told me that defining NULL like that is broken. I agree, but if you don't you end up NULL pointers being in the middle of your address range.
I think most of us can be glad we're not working on Transputers!
I would like to know architectures which violate the assumptions I've
listed below.
I see that Stephen C mentioned PERQ machines, and MSalters mentioned 68000s and PICs.
I'm disappointed that no one else actually answered the question by naming any of the weird and wonderful architectures that have standards-compliant C compilers that don't fit certain unwarranted assumptions.
sizeof(int *) == sizeof(char *) == sizeof(void *) == sizeof(func_ptr
*) ?
Not necessarily. Some examples:
Most compilers for Harvard-architecture 8-bit processors -- PIC and 8051 and M8C -- make sizeof(int *) == sizeof(char *),
but different from the sizeof(func_ptr *).
Some of the very small chips in those families have 256 bytes of RAM (or less) but several kilobytes of PROGMEM (Flash or ROM), so compilers often make sizeof(int *) == sizeof(char *) equal to 1 (a single 8-bit byte), but sizeof(func_ptr *) equal to 2 (two 8-bit bytes).
Compilers for many of the larger chips in those families with a few kilobytes of RAM and 128 or so kilobytes of PROGMEM make sizeof(int *) == sizeof(char *) equal to 2 (two 8-bit bytes), but sizeof(func_ptr *) equal to 3 (three 8-bit bytes).
A few Harvard-architecture chips can store exactly a full 2^16 ("64KByte") of PROGMEM (Flash or ROM), and another 2^16 ("64KByte") of RAM + memory-mapped I/O.
The compilers for such a chip make sizeof(func_ptr *) always be 2 (two bytes);
but often have a way to make the other kinds of pointers sizeof(int *) == sizeof(char *) == sizeof(void *) into a a "long ptr" 3-byte generic pointer that has the extra magic bit that indicates whether that pointer points into RAM or PROGMEM.
(That's the kind of pointer you need to pass to a "print_text_to_the_LCD()" function when you call that function from many different subroutines, sometimes with the address of a variable string in buffer that could be anywhere in RAM, and other times with one of many constant strings that could be anywhere in PROGMEM).
Such compilers often have special keywords ("short" or "near", "long" or "far") to let programmers specifically indicate three different kinds of char pointers in the same program -- constant strings that only need 2 bytes to indicate where in PROGMEM they are located, non-constant strings that only need 2 bytes to indicate where in RAM they are located, and the kind of 3-byte pointers that "print_text_to_the_LCD()" accepts.
Most computers built in the 1950s and 1960s use a 36-bit word length or an 18-bit word length, with an 18-bit (or less) address bus.
I hear that C compilers for such computers often use 9-bit bytes,
with sizeof(int *) == sizeof(func_ptr *) = 2 which gives 18 bits, since all integers and functions have to be word-aligned; but sizeof(char *) == sizeof(void *) == 4 to take advantage of special PDP-10 instructions that store such pointers in a full 36-bit word.
That full 36-bit word includes a 18-bit word address, and a few more bits in the other 18-bits that (among other things) indicate the bit position of the pointed-to character within that word.
The in-memory representation of all pointers for a given architecture
is the same regardless of the data type pointed to?
Not necessarily. Some examples:
On any one of the architectures I mentioned above, pointers come in different sizes. So how could they possibly have "the same" representation?
Some compilers on some systems use "descriptors" to implement character pointers and other kinds of pointers.
Such a descriptor is different for a pointer pointing to the first "char" in a "char big_array[4000]" than for a pointer pointing to the first "char" in a "char small_array[10]", which are arguably different data types, even when the small array happens to start at exactly the same location in memory previously occupied by the big array.
Descriptors allow such machines to catch and trap the buffer overflows that cause such problems on other machines.
The "Low-Fat Pointers" used in the SAFElite and similar "soft processors" have analogous "extra information" about the size of the buffer that the pointer points into. Low-Fat pointers have the same advantage of catching and trapping buffer overflows.
The in-memory representation of a pointer is the same as an integer of
the same bit length as the architecture?
Not necessarily. Some examples:
In "tagged architecture" machines, each word of memory has some bits that indicate whether that word is an integer, or a pointer, or something else.
With such machines, looking at the tag bits would tell you whether that word was an integer or a pointer.
I hear that Nova minicomputers have an "indirection bit" in each word which inspired "indirect threaded code". It sounds like storing an integer clears that bit, while storing a pointer sets that bit.
Multiplication and division of pointer data types are only forbidden
by the compiler. NOTE: Yes, I know this is nonsensical. What I mean is
- is there hardware support to forbid this incorrect usage?
Yes, some hardware doesn't directly support such operations.
As others have already mentioned, the "multiply" instruction in the 68000 and the 6809 only work with (some) "data registers"; they can't be directly applied to values in "address registers".
(It would be pretty easy for a compiler to work around such restrictions -- to MOV those values from an address register to the appropriate data register, and then use MUL).
All pointer values can be casted to a single data type?
Yes.
In order for memcpy() to work right, the C standard mandates that every pointer value of every kind can be cast to a void pointer ("void *").
The compiler is required to make this work, even for architectures that still use segments and offsets.
All pointer values can be casted to a single integer? In other words,
what architectures still make use of segments and offsets?
I'm not sure.
I suspect that all pointer values can be cast to the "size_t" and "ptrdiff_t" integral data types defined in "<stddef.h>".
Incrementing a pointer is equivalent to adding sizeof(the pointed data
type) to the memory address stored by the pointer. If p is an int32*
then p+1 is equal to the memory address 4 bytes after p.
It is unclear what you are asking here.
Q: If I have an array of some kind of structure or primitive data type (for example, a "#include <stdint.h> ... int32_t example_array[1000]; ..."), and I increment a pointer that points into that array (for example, "int32_t p = &example_array[99]; ... p++; ..."), does the pointer now point to the very next consecutive member of that array, which is sizeof(the pointed data type) bytes further along in memory?
A: Yes, the compiler must make the pointer, after incrementing it once, point at the next independent consecutive int32_t in the array, sizeof(the pointed data type) bytes further along in memory, in order to be standards compliant.
Q: So, if p is an int32* , then p+1 is equal to the memory address 4 bytes after p?
A: When sizeof( int32_t ) is actually equal to 4, yes. Otherwise, such as for certain word-addressable machines including some modern DSPs where sizeof( int32_t ) may equal 2 or even 1, then p+1 is equal to the memory address 2 or even 1 "C bytes" after p.
Q: So if I take the pointer, and cast it into an "int" ...
A: One type of "All the world's a VAX heresy".
Q: ... and then cast that "int" back into a pointer ...
A: Another type of "All the world's a VAX heresy".
Q: So if I take the pointer p which is a pointer to an int32_t, and cast it into some integral type that is plenty big enough to contain the pointer, and then add sizeof( int32_t ) to that integral type, and then later cast that integral type back into a pointer -- when I do all that, the resulting pointer is equal to p+1?
Not necessarily.
Lots of DSPs and a few other modern chips have word-oriented addressing, rather than the byte-oriented processing used by 8-bit chips.
Some of the C compilers for such chips cram 2 characters into each word, but it takes 2 such words to hold a int32_t -- so they report that sizeof( int32_t ) is 4.
(I've heard rumors that there's a C compiler for the 24-bit Motorola 56000 that does this).
The compiler is required to arrange things such that doing "p++" with a pointer to an int32_t increments the pointer to the next int32_t value.
There are several ways for the compiler to do that.
One standards-compliant way is to store each pointer to a int32_t as a "native word address".
Because it takes 2 words to hold a single int32_t value, the C compiler compiles "int32_t * p; ... p++" into some assembly language that increments that pointer value by 2.
On the other hand, if that one does "int32_t * p; ... int x = (int)p; x += sizeof( int32_t ); p = (int32_t *)x;", that C compiler for the 56000 will likely compile it to assembly language that increments the pointer value by 4.
I'm most used to pointers being used in a contiguous, virtual memory
space.
Several PIC and 8086 and other systems have non-contiguous RAM --
a few blocks of RAM at addresses that "made the hardware simpler".
With memory-mapped I/O or nothing at all attached to the gaps in address space between those blocks.
It's even more awkward than it sounds.
In some cases -- such as with the bit-banding hardware used to avoid problems caused by read-modify-write -- the exact same bit in RAM can be read or written using 2 or more different addresses.