I've just started to learn C so please be kind.
From what I've read so far regarding pointers:
int * test1; //this is a pointer which is basically an address to the process
//memory and usually has the size of 2 bytes (not necessarily, I know)
float test2; //this is an actual value and usually has the size of 4 bytes,
//being of float type
test2 = 3.0; //this assigns 3 to `test2`
Now, what I don't completely understand:
*test1 = 3; //does this assign 3 at the address
//specified by `pointerValue`?
test1 = 3; //this says that the pointer is basically pointing
//at the 3rd byte in process memory,
//which is somehow useless, since anything could be there
&test1; //this I really don't get,
//is it the pointer to the pointer?
//Meaning, the address at which the pointer address is kept?
//Is it of any use?
Similarly:
*test2; //does this has any sense?
&test2; //is this the address at which the 'test2' value is found?
//If so, it's a pointer, which means that you can have pointers pointing
//both to the heap address space and stack address space.
//I ask because I've always been confused by people who speak about
//pointers only in the heap context.
Great question.
Your first block is correct. A pointer is a variable that holds the address of some data. The type of that pointer tells the code how to interpret the contents of the address being held by that pointer.
The construct:
*test1 = 3
Is called the deferencing of a pointer. That means, you can access the address that the pointer points to and read and write to it like a normal variable. Note:
int *test;
/*
* test is a pointer to an int - (int *)
* *test behaves like an int - (int)
*
* So you can thing of (*test) as a pesudo-variable which has the type 'int'
*/
The above is just a mnemonic device that I use.
It is rare that you ever assign a numeric value to a pointer... maybe if you're developing for a specific environment which has some 'well-known' memory addresses, but at your level, I wouldn't worry to much about that.
Using
*test2
would ultimately result in an error. You'd be trying to deference something that is not a pointer, so you're likely to get some kind of system error as who knows where it is pointing.
&test1 and &test2 are, indeed, pointers to test1 and test2.
Pointers to pointers are very useful and a search of pointer to a pointer will lead you to some resources that are way better than I am.
It looks like you've got the first part right.
An incidental thought: there are various conventions about where to put that * sign. I prefer mine nestled with the variable name, as in int *test1 while others prefer int* test1. I'm not sure how common it is to have it floating in the middle.
Another incidental thought: test2 = 3.0 assigns a floating-point 3 to test2. The same end could be achieved with test2=3, in which case the 3 is implicitly converted from an integer to a floating point number. The convention you have chosen is probably safer in terms of clarity, but is not strictly necessary.
Non-incidentals
*test1=3 does assign 3 to the address specified by test.
test1=3 is a line that has meaning, but which I consider meaningless. We do not know what is at memory location 3, if it is safe to touch it, or even if we are allowed to touch it.
That's why it's handy to use something like
int var=3;
int *pointy=&var;
*pointy=4;
//Now var==4.
The command &var returns the memory location of var and stores it in pointy so that we can later access it with *pointy.
But I could also do something like this:
int var[]={1,2,3};
int *pointy=&var;
int *offset=2;
*(pointy+offset)=4;
//Now var[2]==4.
And this is where you might legitimately see something like test1=3: pointers can be added and subtracted just like numbers, so you can store offsets like this.
&test1 is a pointer to a pointer, but that sounds kind of confusing to me. It's really the address in memory where the value of test1 is stored. And test1 just happens to store as its value the address of another variable. Once you start thinking of pointers in this way (address in memory, value stored there), they become easier to work with... or at least I think so.
I don't know if *test2 has "meaning", per se. In principle, it could have a use in that we might imagine that the * command will take the value of test2 to be some location in memory, and it will return the value it finds there. But since you define test2 as a float, it is difficult to predict where in memory we would end up, setting test2=3 will not move us to the third spot of anything (look up the IEEE754 specification to see why). But I would be surprised if a compiler would allow such thing.
Let's look at another quick example:
int var=3;
int pointy1=&var;
int pointy2=&pointy1;
*pointy1=4; //Now var==4
**pointy2=5; //Now var==5
So you see that you can chain pointers together like this, as many in a row as you'd like. This might show up if you had an array of pointers which was filled with the addresses of many structures you'd created from dynamic memory, and those structures contained pointers to dynamically allocated things themselves. When the time comes to use a pointer to a pointer, you'll probably know it. For now, don't worry too much about them.
First let's add some confusion: the word "pointer" can refer to either a variable (or object) with a pointer type, or an expression with the pointer type. In most cases, when people talk about "pointers" they mean pointer variables.
A pointer can (must) point to a thing (An "object" in standards parlance). It can only point to the right kind of thing; a pointer to int is not supposed to point to a float object. A pointer can also be NULL; in that case there is no thing to point to.
A pointertype is also a type, and a pointer object is also an object. So it is allowable to construct a pointer to pointer: the pointer-to-pointer just stores the addres of the pointer object.
What a pointer can not be:
It cannot point to a value: p = &4; is impossible. 4 is a literal value, which is not stored in an object, and thus has no address.
the same goes for expressions: p = &(1+4); is impossible, because the expression "1+4" does not have a location.
the same goes for return value p = &sin(pi); is impossible; the return value is not an object and thus has no address.
variables marked as "register" (almost distinct now) cannot have an address.
you cannot take the address of a bitfield, basically because these can be smaller than character (or have a finer granularity), hence it would be possible that different bitmasks would have the same address.
There are some "exceptions" to the above skeletton (void pointers, casting, pointing one element beyond an array object) but for clarity these should be seen as refinements/amendments, IMHO.
Related
As I understand it, all of the cases where C has to handle an address involve the use of a pointer. For example, the & operand creates a pointer to the program, instead of just giving the bare address as data (i.e it never gives the address without using a pointer first):
scanf("%d", &foo)
Or when using the & operand
int i; //a variable
int *p; //a variable that store adress
p = &i; //The & operator returns a pointer to its operand, and equals p to that pointer.
My question is: Is there a reason why C programs always have to use a pointer to manage addresses? Is there a case where C can handle a bare address (the numerical value of the address) on its own or with another method? Or is that completely impossible? (Being because of system architecture, memory allocation changing during and in each runtime, etc). And finally, would that be useful being that addresses change because of memory management? If that was the case, it would be a reason why pointers are always needed.
I'm trying to figure out if the use pointers is a must in C standardized languages. Not because I want to use something else, but because I want to know for sure that the only way to use addresses is with pointers, and just forget about everything else.
Edit: Since part of the question was answered in the comments of Eric Postpischil, Michał Marszałek, user3386109, Mike Holt and Gecko; I'll group those bits here: Yes, using bare adresses bear little to no use because of different factors (Pointers allow a number of operations, adresses may change each time the program is run, etc). As Michał Marszałek pointed out (No pun intended) scanf() uses a pointer because C can only work with copies, so a pointer is needed to change the variable used. i.e
int foo;
scanf("%d", foo) //Does nothing, since value can't be changed
scanf("%d", &foo) //Now foo can be changed, since we use it's address.
Finally, as Gecko mentioned, pointers are there to represent indirection, so that the compiler can make the difference between data and address.
John Bode covers most of those topics in it's answer, so I'll mark that one.
A pointer is an address (or, more properly, it’s an abstraction of an address). Pointers are how we deal with address values in C.
Outside of a few domains, a “bare address” value simply isn’t useful on its own. We’re less interested in the address than the object at that address. C requires us to use pointers in two situations:
When we want a function to write to a parameter
When we need to track dynamically allocated memory
In these cases, we don’t really care what the address value actually is; we just need it to access the object we’re interested in.
Yes, in the embedded world specific address values are meaningful. But you still use pointers to access those locations. Like I said above, a pointer is an address for our purposes.
C allows you to convert pointers to integers. The <stdint.h> header provides a uintptr_t type with the property that any pointer to void can be converted to uintptr_t and back, and the result will compare equal to the original pointer.
Per C 2018 6.3.2.3 6, the result of converting a pointer to an integer is implementation-defined. Non-normative note 69 says “The mapping functions for converting a pointer to an integer or an integer to a pointer are intended to be consistent with the addressing structure of the execution environment.”
Thus, on a machine where addresses are a simple numbering scheme, converting a pointer to a uintptr_t ought to give you the natural machine address, even though the standard does not require it. There are, however, environments where addresses are more complicated, and the result of converting a pointer to an integer may not be straightforward.
int i; //a variable
int *p; //a variable that store adres
i = 10; //now i is set to 10
p = &i; //now p is set to i address
*p = 20; //we set to 20 the given address
int tab[10]; // a table
p = tab; //set address
p++; //operate on address and move it to next element tab[1]
We can operate on address by pointers move forward or backwards. We can set and read from given address.
In C if we want get return values from functions we must use pointers. Or use return value from functions, but that way we can only get one value.
In C we don't have references therefore we must use pointers.
void fun(int j){
j = 10;
}
void fun2(int *j){
*j = 10;
}
int i;
i = 5; // now I is set to 5
fun(i);
//printf i will print 5
fun2(&i);
//printf I will print 10
I'm learning about pointers and have been told this: "The purpose of pointers is to allow you to manually, directly access a block of memory."
Say I have int var = 5;. Can't I use the variable 'var' to access the block of memory where the value 5 is stored, since I can change the value of the variable whenever I want var = 6;? Do I really need a pointer when I can access any variable's value just by using its variable, instead of using a pointer that points to the address where the value is stored?
"The purpose of pointers is to allow you to manually, directly access a block of memory."
This is not always true. Consider
*(int*)(0x1234) = some_value;
this is "direct" memory access. Though
int a = some_value, *ptr = &a;
*ptr = some_other_value;
you are now accessing a indirectly.
Can't I use the variable 'var' to access the block of memory where the
value 5 is stored, since I can change the value of the variable
whenever I want var = 6; ?
Surely; but the semantics is different.
Do I really need a pointer when I can access any variable's value just by using its variable, instead of using a pointer that points to the address where the value is stored?
No, you don't. Consider the first example: within the scope where a has been declared, modifying its value through ptr is rather pointless! However, what if you are not within the scope of a? That is
void foo(int x)
{
x = 5;
}
int main(void)
{
int x = 10;
foo(x);
}
In foo, when you do x = 5, there is an ambiguity: do you want to modify foo::x or main::x? In the latter case that has to be "requested" explicitly and the fact that happens through pointers -or, better, through indirection- is a coincidence and a language choice. Other languages have others.
Pointer types have some traits that make them really useful:
It's guaranteed that a pointer will be so large that it can hold any address that is supported by the architecture (on x86, that is 32 bits a.k.a. 4 bytes, and an x64 64 bits a.k.a. 8 bytes).
Dereferencing and indexing the memory is done per object, not per byte.
int buffer[10];
char*x = buffer;
int*y = (int*)buffer;
That way, x[1] isn't y[1]
Both is not guaranteed if you use simple ints to hold your values. The first trait is at least guaranteed by uintptr_t (not by size_t though, although most of the time they have the same size - except that size_t can be 2 bytes in size on systems with segmented memory layout, while uintptr_t is still 4 bytes in size).
While using ints might work at first, you always:
have to turn the value into a pointer
have to dereference the pointer
and have to make sure that you don't go beyond certain values for your "pointer". For a 16 bit int, you cannot go beyond 0xFFFF, for 32 bit it's 0xFFFF FFFF - once you do, your pointer might overflow without you noticing it until it's too late.
That is also the reason why linked lists and pointers to incomplete types work - the compiler already knows the size of the pointers you are going to you, and just allocates memory for them. All pointers have the same size (4 or 8 bytes on 32-bit/64-bit architectures) - the type that you assign them just tells the compiler how to dereference the value. char*s take up the same space as void*s, but you cannot dereference void*s. The compiler won't let you.
Also, if you are just dealing with simple integers, there's a good chance that you will slow down your program significantly do to something called "aliasing", which basically forces the compiler to read the value of a given address all the time. Memory accesses are slow though, so you want to optimized these memory accesses out.
You can compare a memory address to a street address:
If you order something, you tell the shop your address, so that they can send you what you bought.
If you don't want to use your address, you have to send them your house, such that they can place the parcel inside.
Later they return your house to you. This is a bit more cumbersome than using the address!
If you're not at home, the parcel can be delivered to your neighbor if they have your address, but this is not possible if
you sent them your house instead.
The same is true for pointers: They are small and can be transported easily, while the object they point to
might be large, and less easily transportable.
With pointer arithmetics, pointers can also be used to access other objects than the one they originally pointed to.
You call a function from main() or from another function, the function you called can only return 1 value.
Let say you want 3 values changed, you pass them to the called function as pointers. That way you don't have to use global values.
One possible advantage is that it can make it easier to have one function modify a variable that will be used by many other functions. Without pointers, your best option is for the modifying function to return a value to the caller and then to pass this value to the other functions. That can lead to a lot of passing around. Instead, you can give the modifying function a pointer where it stores its output, and all the other functions directly access that memory address. Kind of like global variables.
I have a
LS_Led* LS_vol_leds[10];
declared in one C module, and the proper externs in the other modules that access it.
In func1() I have this line:
/* Debug */
LS_Led led = *(LS_vol_leds[0]);
And it does not cause an exception. Then
I call func2() in another C module (right after above line), and do the same line, namely:
/* Debug */
LS_Led led = *(LS_vol_leds[0]);`
first thing, and exception thrown!!!
I don't think I have the powers to debug this one on my own.
Before anything LS_vol_leds is initialized in func1() with:
LS_vol_leds[0] = &led3;
LS_vol_leds[1] = &led4;
LS_vol_leds[2] = &led5;
LS_vol_leds[3] = &led6;
LS_vol_leds[4] = &led7;
LS_vol_leds[5] = &led8;
LS_vol_leds[6] = &led9;
LS_vol_leds[7] = &led10;
LS_vol_leds[8] = &led11;
LS_vol_leds[9] = &led12;
My externs look like
extern LS_Led** LS_vol_leds;
So does that lead to disaster and I how do I prevent disaster?
Thanks.
This leads to disaster:
extern LS_Led** LS_vol_leds;
You should try this instead:
extern LS_Led *LS_vol_leds[];
If you really want to know why, you should read Expert C Programming - Deep C Secrets, by Peter Van Der Linden (amazing book!), especially chapter 4, but the quick answer is that this is one of those corner cases where pointers and arrays are not interchangeable: a pointer is a variable which holds the address of another one, whereas an array name is an address. extern LS_Led** LS_vol_leds; is lying to the compiler and generating the wrong code to access LS_vol_leds[i].
With this:
extern LS_Led** LS_vol_leds;
The compiler will believe that LS_vol_leds is a pointer, and thus, LS_vol_leds[i] involves reading the value stored in the memory location that is responsible for LS_vol_leds, use that as an address, and then scale i accordingly to get the offset.
However, since LS_vol_leds is an array and not a pointer, the compiler should instead pick the address of LS_vol_leds directly. In other words: what is happening is that your original extern causes the compiler to dereference LS_vol_leds[0] because it believes that LS_vol_leds[0] holds the address of the pointed-to object.
UPDATE: Fun fact - the back cover of the book talks about this specific case:
So that's why extern char *cp isn't the same as extern char cp[]. I
knew that it didn't work despite their superficial equivalence, but I
didn't know why. [...]
UPDATE2: Ok, since you asked, let's dig deeper. Consider a program split into two files, file1.c and file2.c. Its contents are:
file1.c
#define BUFFER_SIZE 1024
char cp[BUFFER_SIZE];
/* Lots of code using cp[i] */
file2.c
extern char *cp;
/* Code using cp[i] */
The moment you try to assing to cp[i] or use cp[i] in file2.c will most likely crash your code. This is deeply tight into the mechanics of C and the code that the compiler generates for array-based accesses and pointer-based accesses.
When you have a pointer, you must think of it as a variable. A pointer is a variable like an int, float or something similar, but instead of storing an integer or a float, it stores a memory address - the address of another object.
Note that variables have addresses. When you have something like:
int a;
Then you know that a is the name for an integer object. When you assign to a, the compiler emits code that writes into whatever address is associated with a.
Now consider you have:
char *p;
What happens when you access *p? Remember - a pointer is a variable. This means that the memory address associated with p holds an address - namely, an address holding a character. When you assign to p (i.e., make it point to somewhere else), then the compiler grabs the address of p and writes a new address (the one you provide it) into that location.
For example, if p lives at 0x27, it means that reading memory location 0x27 yields the address of the object pointed to by p. So, if you use *p in the right hand side of an assignment, the steps to get the value of *p are:
Read the contents of 0x27 - say it's 0x80 - this is the value of the pointer, or, equivalently, the address of the pointed-to object
Read the contents of 0x80 - this finally gives you *p.
What if p is an array? If p is an array, then the variable p itself represents the array. By convention, the address representing an array is the address of its first element. If the compiler chooses to store the array in address 0x59, it means that the first element of p lives at 0x59. So when you read p[0] (or *p), the generated code is simpler: the compiler knows that the variable p is an array, and the address of an array is the address of the first element, so p[0] is the same as reading 0x59. Compare this to the case for which p is a pointer.
If you lie to the compiler, and tell it you have a pointer instead of an array, the compiler will (wrongly) generate code that does what I showed for the pointer case. You're basically telling it that 0x59 is not the address of an array, it's the address of a pointer. So, reading p[i] will cause it to use the pointer version:
Read the contents of 0x59 - note that, in reality, this is p[0]
Use that as an address, and read its contents.
So, what happens is that the compiler thinks that p[0] is an address, and will try to use it as such.
Why is this a corner case? Why don't I have to worry about this when passing arrays to functions?
Because what is really happening is that the compiler manages it for you. Yes, when you pass an array to a function, a pointer to the first element is passed, and inside the called function you have no way to know if it is a "real" array or a pointer. However, the address passed into the function is different depending on whether you're passing a real array or a pointer. If you're passing a real array, the pointer you get is the address of the first element of the array (in other words: the compiler immediately grabs the address associated to the array variable from the symbol table). If you're passing a pointer, the compiler passes the address that is stored in the address associated with that variable (and that variable happens to be the pointer), that is, it does exactly those 2 steps mentioned before for pointer-based access. Again, note that we're discussing the value of the pointer here. You must keep this separated from the address of the pointer itself (the address where the address of the pointed-to object is stored).
That's why you don't see a difference. In most situations, arrays are passed around as function arguments, and this rarely raises problems. But sometimes, with some corner cases (like yours), if you don't really know what is happening down there, well.. then it will be a wild ride.
Personal advice: read the book, it's totally worth it.
Okay, currently I am writing a kernel for the sake of my resume. While writing my memory management unit I have hit a brick wall.
int address = (int)malloc(sizeof(Test))
consoleWriteString("Variable Address:\n");
consoleWriteInteger(address);
char* f = (void*)address;
consoleWriteString("\nVariable Address:\n");
consoleWriteInteger((int)&f); // Should print off the same as above
Logically the output should be the same for both. Somewhere somthing has gone wrong though. as my output is the following.
Variable Address: 47167
Variable Address: 1065908
After a long period of testing and debugging I finally gave in and decided to ask stack overflow. Also if you spot any syntax errors ignore them. By the way this is all in C, and all functions are custom, including malloc, but I have determined that the error does not lie in that functon, or any other for a fact. I believe this is just me being stupid about pointers and casting but don't laugh at me when it was somthing super simple that I missed.
Thanks yall
&f is the address of f, not the address contained in it! f is on the stack. Its value (the address you first printed) is pointing to the allocated memory.
Think of it this way: You allocate room in memory for some stuff. This memory region has an address. You put the address in a pointer (f), so that it points to that region. But f itself needs to be somewhere in memory in order to hold the value of that address. In this case, f is on the stack, and &f gets the address of f (the container of the original address), not the address that f contains.
As an aside, be very careful casting addresses to int (and back!), since int might not be large enough to hold an address (e.g. on x86-64, depending on your compiler). I believe the correct type to use when you want to use an address as an integer is uintptr_t in stdint.h.
The value of f (which happens to be a pointer) is the same as address (which is also a pointer, but of a different type) - this is what you do in the line
char* f = (void*)address;
But then you print the address of f:
consoleWriteInteger((int)&f);
And that is not the same thing as the value of f... change that line to
consoleWriteInteger((int)f);
and you should be all set.
the first print out is an int, although malloc returns an address, The second is an address casted to an int. do f instead & gets you the address of a value while * dereferences a pointer getting you the value of what is being pointed to.
Yesterday while I was coding in C, my friend asked me pointing to a variable is it pointer or a variable ? I stucked up for a while. I didnt find an aswer to it , I just have to go back and search it and tell him.But I was thinking is there any function to differentiate them.
Can we differentiate a variable against a pointer variable
int a;
sizeof(a); // gives 2 bytes
int *b;
sizeof(b); // gives 2 bytes
// if we use sizeof() we get same answer and we cant say which is pointer
// and which is a variable
Is there a way to find out a variable is a normal variable or a pointer? I mean can someone say that it is a pointer or a variable after looking at your variable that you have declared at the beginning and then going down 1000 lines of your code?
After the comment
I wanted to say explicitly it's a 16 bit system architecture.
First, the question "Is it a pointer or a variable" doesn't make much sense. A pointer variable is a variable, just as an integer variable, or an array variable, is a variable.
So the real question is whether something is a pointer or not.
No, there's no function that can tell you whether something is a pointer or not. And if you think about it, in a statically typed language like C, there can't be. Functions take arguments of certain specified types. You can't pass a variable to a function unless the type (pointer or otherwise) is correct in the first place.
You mean differentiate them at run time without seeing the code? No, you can't. Pointers are variables that hold memory address. You can't check it at run time. That means, there is no such function isPointer(n) that will return true/false based on parameter n.
You can deduce the type from the use.
For example:
char* c;
...
c[0] = 'a';
*c = 'a';
Indexing and dereferencing would let you know it's a pointer to something (or it's an array if defined as char c[SOME_POSITIVE_NUMBER];).
Also, things like memset(c,...), memcpy(c,...) will suggest that c is a pointer (array).
OTOH, you can't normally do with pointers most of arithmetic, so, if you see something like
x = c * 2;
y = 3 / c;
z = c << 1;
w = 1 & c;
then c is not a pointer (array).
Three things:
What platform are you using where sizeof(int) returns 2? Seriously
Pointers are types. A pointer to an int is a type, just like an int is. The sizes of a type and a pointer to that type are sometimes equal but not directly related; for instance, a pointer to a double (on my machine, at least) has size 4 bytes while a double has size 8 bytes. sizeof() would be a very poor test, even if there was a situation where such a test would be appropriate (there isn't).
C is a strictly typed language, and your question doesn't really make sense in that context. As the programmer, you know exactly what a is and you will use it as such.
If you'd like to be able to tell whether a variable is a pointer or not when you see it in the source code, but without going back to look at the declaration, a common approach is to indicate it in the way you name your variables. For example, you might put a 'p' at the beginning of the names of pointers:
int *pValue; /* starts with 'p' for 'pointer' */
int iOther; /* 'i' for 'integer' */
...or even:
int *piSomething; /* 'pi' for 'Pointer to Integer' */
This makes it easy to tell the types when you see the variable in your code. Some people use quite a range of prefixes, to distinguish quite a range of types.
Try looking up "Hungarian notation" for examples.
no , you can't.
and what is the usage, as each time u run the code the pointer address will be different ?? however u can subtract two pointers and also can get the memory address value of any pointer.