Know if const qualifier is used - c

Is there any way in C to find if a variable has the const qualifier? Or if it's stored in the .rodata section?
For example, if I have this function:
void foo(char* myString) {...}
different actions should be taken in these two different function calls:
char str[] = "abc";
foo(str);
foo("def");
In the first case I can modify the string, in the second one no.

Not in standard C, i.e. not portably.
myString is just a char* in foo, all other information is lost. Whatever you feed into the function is automatically converted to char*.
And C does not know about ".rodata".
Depending on your platform you could check the address in myString (if you know your address ranges).

You can't differ them using the language alone. In other words, this is not possible without recurring to features specific to the compiler you're using, which is likely not to be portable. A few important remarks though:
In the first case you COULD modify the string, but you MUST NOT. If you want a mutable string, use initialization instead of assignment.
char *str1 = "abc"; // NOT OK, should be const char *
const char *str2 = "abc"; // OK, but not mutable
char str3[] = "abc"; // OK, using initialization, you can change its contents

#include<stdio.h>
void foo(char *mystr)
{
int a;
/*code goes here*/
#ifdef CHECK
int local_var;
printf(" strings address %p\n",mystr);
printf("local variables address %p \n",&local_var);
puts("");
puts("");
#endif
return;
}
int main()
{
char a[]="hello";
char *b="hello";
foo(a);
foo(b);
foo("hello");
}
On compiling with gcc -DCHECK prog_name.c and executing on my linux machine the following output comes...
strings address 0xbfdcacf6
local variables address 0xbfdcacc8
strings address 0x8048583
local variables address 0xbfdcacc8
strings address 0x8048583
local variables address 0xbfdcacc8
for first case when string is defined and initialized in the "proper c way for mutable strings" the difference between the addresses is 0x2E.(5 bytes).
in the second case when string is defined as char *p="hello" the differences in addresses is
0xB7D82745.Thats bigger than the size of my stack.so i am pretty sure the string is not on the stack.Hence the only place where you can find it is .rodata section.
The third one is similar case
PS:As mentioned above this isn't portable but the original question hardly leaves any scope for portability by mentioning .rodata :)

GCC provides the __builtin_constant_p builtin function, which enables you to determine whether an expression is constant or not at compile-time:
Built-in Function: int __builtin_constant_p (exp)
You can use the built-in function __builtin_constant_p to determine if a value is known to be constant at compile-time and hence that GCC can perform constant-folding on expressions involving that value. The argument of the function is the value to test. The function returns the integer 1 if the argument is known to be a compile-time constant and 0 if it is not known to be a compile-time constant. A return of 0 does not indicate that the value is not a constant, but merely that GCC cannot prove it is a constant with the specified value of the `-O' option.
So I guess you should rewrite your foo function as a macro in such a case:
#define foo(x) \
(__builtin_constant_p(x) ? foo_on_const(x) : foo_on_var(x))
foo("abc") would expand to foo_on_const("abc") and foo(str) would expand to foo_on_var(str).

Related

Why are strings in C declared with 'const'?

For example, why not:
char *s= "example";
instead of:
const char *s= "example";
I understand that const makes it unchangeable, but why do I receive an error when compiling the first?
Additionally, how does the concept apply to
int * x;
vs
const int *x;
I see the second used a lot more, is it good practice to use "cons int *"?
There's no requirement to use const, but it's a good idea.
In C, a string literal is an expression of type char[N], where N is the length of the string plus 1 (for the terminating '\0' null character). But attempting to modify the array that corresponds to the string literal has undefined behavior. Many compilers arrange for that array to be stored in read-only memory (not physical ROM, but memory that's marked read-only by the operating system). (An array expression is, in most contexts converted to a pointer expression referring to the initial element of the array object.)
It would have made more sense to make string literals const, but the const keyword did not exist in old versions of C, and it would have broken existing code. (C++ did make string literals const).
This:
char *s= "example"; /* not recommended */
is actually perfectly valid in C, but it's potentially dangerous. If, after this declaration, you do:
s[0] = 'E';
then you're attempting to modify the string literal, and the behavior is undefined.
This:
const char *s= "example"; /* recommended */
is also valid; the char* value that results from evaluating the string literal is safely and quietly converted to const char*. And it's generally better than the first version because it lets the compiler warn you if you attempt to modify the string literal (it's better to catch errors at compile time than at run time).
If you get an error on your first example, then it's likely that you're inadvertently compiling your code as C++ rather than as C -- or that you're using gcc's -Wwrite-strings option or something similar. (-Wwrite-strings makes string literals const; it can improve safety, but it can also cause gcc to reject, or at least warn about, valid C code.)
With Visual Studio 2015 at warning level 4, this compiles and runs whether compiled as C or C++:
#include <stdio.h>
char *s1= "example\n";
const char *s2= "example\n";
int main(int argc, char **argv)
{
printf(s1); // prints "example"
s1[2] = 'x';
printf(s1); // prints "exxmple"
printf(s2);
return 0;
}
If I add this line, it will fail to compile as C or C++ with every compiler I know of:
s2[2] = 'x'; // produces compile error
This is the error the const keyword is designed to avoid. It simply tells the compiler not to allow assignments to the object pointed to.
It doesn't matter if your pointer points to char or int or anything else. The const keyword has the same effect on all pointers, and that's to make it impossible (well, very hard) to assign to the thing declared const.
A string literal used as a value compiles to an array of char that should not be modified. Attempting to modify it invokes undefined behavior. For historical reasons of backward compatibility, its type is char [] although is really should be const char []. You can enable extra compiler warnings to change this and instruct the compiler to consider such strings to be const.

C Aligning string literals for a specific use case

I'm trying to align string literals in a specific way, as how I'm using it in my code is fairly specific. I don't want to have to assign it to a variable, for instance many of my functions are using it as a direct argument. And I want it to work both in local scope or global scope.
Usage example:
char *str = ALIGNED_STRING("blah"); //what I want
foo(ALIGNED_STRING("blah")); //what I want
_Alignas(16) char str[] = "blah"; //not what I want (but would correctly align the string)
The ideal solution would be (_Alignas(16) char[]){ "blah" } or a worser case using the GCC/Clang compiler extensions for alignment (__attribute__((alignment(16))) char[]){ "blah" }, but neither works (they're ignored and the default alignment for the type is used).
So my next thought was to align it myself, and then my functions that use the string could then fix it up correctly. e.g. #define ALIGNED_STRING(str) (char*)(((uintptr_t)(char[]){ "xxxxxxxxxxxxxxx" str } + 16 - 1) & ~(16 - 1)) (where the string containing 'x' would represent data needed to understand where the real string can be found, that's easy but just for the example assume the 'x' is fine). Now that works fine in local scope, but fails in the global scope. Since the compiler complains about it not being a compile-time constant (error: initializer element is not a compile-time constant); I would've thought it would work but it seems only addition and subtraction are valid operations on the pointer at compile-time.
So I'm wondering if there's anyway to achieve what I want to do? At the moment I'm just using the latter example (padding and manually aligning) and avoiding to use it in the global scope (but I would really want to). And the best solution would avoid needing to make runtime adjustments (like using the alignment qualifier would), but that doesn't seem possible unless I apply it to a variable (but as mentioned that's not what I want to do).
Was able to get close to OP's need with a compound literal. (C99)
#include <stdio.h>
#include <stddef.h>
void bar(const char *s) {
printf("%p %s\n", (void*)s, s);
}
// v-- compound literal --------------------------v
#define ALIGNED_STRING(S) (struct { _Alignas(16) char s[sizeof S]; }){ S }.s
int main() {
char s[] = "12";
bar(s);
char t[] = "34";
bar(t);
bar(ALIGNED_STRING("asdfas"));
char *u = ALIGNED_STRING("agsdas");
bar(u);
}
Output
0x28cc2d 12
0x28cc2a 34
0x28cc30 asdfas // 16 Aligned
0x28cc20 agsdas // 16 Aligned

Perform special kind of initialization of string in C

As you know in C, we can initialize string variables like this:
char text[1024] =
"Hello "
"World";
But what if I have a function that returns the word "World"?
char text[1024] =
"Hello "
World();
It seems to me that's not possible in C.
Please confirm.
What you want is not possible.
The L-value to the assigment operator needs to be modifyable, which an array isn't.
From the C11-Standard:
6.5.16/2
An assignment operator shall have a modifiable lvalue as its left operand.
The only exception to this is during initialisation when using literals as R-value:
char text[1024] = "Hello ""World";
From the C11-Standard:
6.7.9/14
An array of character type may be initialized by a character string literal or UTF−8 string
literal, optionally enclosed in braces. Successive bytes of the string literal (including the
terminating null character if there is room or if the array is of unknown size) initialize the
elements of the array.
If World() is something that always returns "World", then define it as a macro:
#define World "World"
And then do:
char text[1024] =
"Hello "
World; //Without parentheses
EDIT
String concatenation in the way you expect to do is made by the C preprocessor.You are actually looking for a runtime concatenation of two strings, which can be performed in multiple ways. The simplest one is achieved by strcat function, but the initialization should be performed explicitly by a function:
char text[1024];
void init_text() {
strcpy(text, "Hello ");
strcat(text, World()); //World() defined somewhere else
}
Alternative using sprintf :
void init_text() {
sprintf(text, "Hello %s", World());
}
Then in the main function, call init_text() at the beginning:
int main() {
init_text();
...
}
It is not possible in standard C to initialize something with some runtime specific behavior. So the standard portable way is to initialize the data by calling a function at the beginning of main, as answered by Claudix.
However, if you are using some recent GCC compiler (or Clang/LLVM) you could otherwise, on some systems (including Linux and probably other POSIX systems), use some constructor attribute on function. So you would declare:
static void init_text(void) __attribute__((constructor));
and define init_text like in Claudix's answer without having to call it in your main : since it has the constructor attribute, it would be called magically before main, or during dlopen(3) if it appears inside a dynamically linked plugin or library.
A more portable trick might be to have a function returning that text which will initialize it during its first call. So instead of using text you would call get_my_text() everywhere (perhaps by putting #define text get_my_text() in a header, but I don't recommend doing so for readability reasons, so replace every occurrence of text by get_my_text() ...), and define it as:
const char*get_my_text() {
static char textbuf[1024];
if (textbuf[0]) {
// already initialized, so return it
return textbuf;
};
snprintf(textbuf, sizeof(textbuf), "Hello %s", World());
return textbuf;
}
Beware that such a trick is not reliable in multi-threaded programs: two threads might run get_my_text exactly at the same time, and you have a data race. In a multi-threaded app use e.g. pthread_once
You could even define get_my_text as a static inline function in your header file.
PS Always prefer snprintf(3) to sprintf to avoid buffer overflows. Also notice that in standard C++ any static or global data with some given constructor is initialized before main ... hence the name of GCC function attribute...

Difference between literal constant & normal variable?

I am learning C programming, and my book says that unlike variables, constants cannot be changed during the program's execution. And that their are two types of constants Literal and Symbolic. I think that I understand Symbolic pretty well. But Literal Constants are confusing me.
The example it gave me for was
int count = 20;
I wrote this simple program, and I could change the value of the Literal Constant.
/* Demonstrates variables and constants */
#include <stdio.h>
/* Trying to figure out if literal constants are any different from variables */
int testing = 22;
int main( void )
{
/* Print testing before changing the value */
printf("\nYour int testing has a value of %d", testing);
/* Try to change the value of testing */
testing = 212345;
/* Print testing after changing the value */
printf("\nYour int testing has a value of %d", testing);
return 0;
}
It outputted this:
Your int testing has a value of 22
Your int testing has a value of 212345
RUN SUCCESSFUL (total time: 32ms)
Can someone explain how this happens, am I declaring it wrong? Or is there any difference between normal variable and literal constants?
-Thanks
The literal constant is the 20. You can change the value of count, but you cannot change the value of 20 to be, for example, 19.
(As some trivia, there are versions of FORTRAN where you could do exactly this, so it's not meaningless to talk about)
The literal in this case is the number 22 (and later the number 212345). You assign this literal to a variable testing. This variable can be changed.
This is a little trickier when it comes to strings and string literals. If you have a pointer to a string literal, you can change the actual pointer to point to some other string, but you can change what the original pointer points to.
For example:
const char *string_pointer = "Foobar";
With the above definition, you can not do e.g. string_pointer[0] = 'L';, as that is an attempt to modify the literal string that the pointer points to.
Depending on context, a symbolic constant can be either a "variable" declared as const or a preprocessor macro.
A literal constant in C is just any number itself, e.g. your 20. You cannot change it by doing, say, 20 = 50;. That is illegal.
There are also constant variables, e.g. const int blah = 42;. You cannot change this blah by doing something like blah = 100;.
But if you have a normal, non-constant variable, e.g. int foo = 123;, you can change it, e.g. foo = 456;.

Pointer to literal value

Suppose I have a constant defined in a header file
#define THIS_CONST 'A'
I want to write this constant to a stream. I do something like:
char c = THIS_CONST;
write(fd, &c, sizeof(c))
However, what I would like to do for brevity and clarity is:
write(fd, &THIS_CONST, sizeof(char)); // error
// lvalue required as unary ‘&’ operand
Does anyone know of any macro/other trick for obtaining a pointer to a literal? I would like something which can be used like this:
write(fd, PTR_TO(THIS_CONST), sizeof(char))
Note: I realise I could declare my constants as static const variables, but then I can't use them in switch/case statements. i.e.
static const char THIS_CONST = 'A'
...
switch(c) {
case THIS_CONST: // error - case label does not reduce to an integer constant
...
}
Unless there is a way to use a const variable in a case label?
There is no way to do this directly in C89. You would have to use a set of macros to create such an expression.
In C99, it is allowed to declare struct-or-union literals, and initializers to scalars can be written using a similar syntax. Therefore, there is one way to achieve the desired effect:
#include <stdio.h>
void f(const int *i) {
printf("%i\n", *i);
}
int main(void) {
f(&(int){1});
return 0;
}
These answers are all outdated, and apart from a comment nobody refers to recent language updates.
On a C99-C11-C17 compiler using a compound literal, http://en.cppreference.com/w/c/language/compound_literal, is possible to create
a pointer to a nameless constant, as in:
int *p = &((int){10});
The only way you can obtain a pointer is by putting the literal into a variable (your first code example). You can then use the variable with write() and the literal in switch.
C simply does not allow the address of character literals like 'A'. For what it's worth, the type of character literals in C is int (char in C++ but this question is tagged C). 'A' would have an implementation defined value (such as 65 on ASCII systems). Taking the address of a value doesn't make any sense and is not possible.
Now, of course you may take the address of other kinds of literals such as string literals, for example the following is okay:
write(fd, "potato", sizeof "potato");
This is because the string literal "potato" is an array, and its value is a pointer to the 'p' at the start.
To elaborate/clarify, you may only take the address of objects. ie, the & (address-of) operator requires an object, not a value.
And to answer the other question that I missed, C doesn't allow non-constant case labels, and this includes variables declared const.
Since calling write() to write a single character to a file descriptor is almost certainly a performance killer, you probably want to just do fputc( THIS_CONST, stream ).
#define THIS_CONST 'a'
Is just a macro. The compiler basically just inserts 'a' everywhere you use THIS_CONST.
You could try:
const char THIS_CONST = 'a';
But I suspect that will not work wither (don't have a c-compiler handy to try it out on, and it has been quite a few years since I've written c code).
just use a string constant, which is a pointer to a character, and then only write 1 byte:
#define MY_CONST_STRING "A"
write(fd, MY_CONST_STRING, 1);
Note that the '\0' byte at the end of the string is not written.
You can do this for all sorts of constant values, just use the appropriate hex code string, e.g.
#define MY_CONST_STRING "\x41"
will give also the character 'A'. For multiple-byte stuff, take care that you use the correct endianness.
Let's say you want to have a pointer to a INT_MAX, which is e.g. 0x7FFFFFFF on a 32 bit system. Then you can do the following:
#define PTR_TO_INT_MAX "\xFF\xFF\xFF\x7F"
You can see that this works by passing it as a dereferenced pointer to printf:
printf ("max int value = %d\n", *(int*)PTR_TO_INT_MAX);
which should print 2147483647.
For chars, you may use extra global static variables. Maybe something like:
#define THIS_CONST 'a'
static char tmp;
#define PTR_TO(X) ((tmp = X),&tmp)
write(fd,PTR_TO(THIS_CONST),sizeof(char));
There's no reason the compiler has to put the literal into any memory location, so your question doesn't make sense. For example a statement like
int a;
a = 10;
would probably just be directly translated into "put a value ten into a register". In the assembly language output of the compiler, the value ten itself never even exists as something in memory which could be pointed at, except as part of the actual program text.
You can't take a pointer to it.
If you really want a macro to get a pointer,
#include <stdio.h>
static char getapointer[1];
#define GETAPOINTER(x) &getapointer, getapointer[0] = x
int main ()
{
printf ("%d\n",GETAPOINTER('A'));
}
I can see what you're trying to do here, but you're trying to use two fundamentally different things here. The crux of the matter is that case statements need to use values which are present at compile time, but pointers to data in memory are available only at run time.
When you do this:
#define THIS_CONST 'A'
char c = THIS_CONST;
write(fd, &c, sizeof(c))
you are doing two things. You are making the macro THIS_CONST available to the rest of the code at compile time, and you are creating a new char at runtime which is initialised to this value. At the point at which the line write(fd, &c, sizeof(c)) is executed, the concept of THIS_CONST no longer exists, so you have correctly identified that you can create a pointer to c, but not a pointer to THIS_CONST.
Now, when you do this:
static const char THIS_CONST = 'A';
switch(c) {
case THIS_CONST: // error - case label does not reduce to an integer constant
...
}
you are writing code where the value of the case statement needs to be evaluated at compile time. However, in this case, you have specified THIS_CONST in a way where it is a variable, and therefore its value is available only at runtime despite you "knowing" that it is going to have a particular value. Certainly, other languages allow different things to happen with case statements, but those are the rules with C.
Here's what I'd suggest:
1) Don't call a variable THIS_CONST. There's no technical reason not to, but convention suggests that this is a compile-time macro, and you don't want to confuse your readers.
2) If you want the same values to be available at compile time and runtime, find a suitable way of mapping compile-time macros into run-time variables. This may well be as simple as:
#define CONST_STAR '*'
#define CONST_NEWLINE '\n'
static const char c_star CONST_STAR;
static const char c_newline CONST_NEWLINE;
Then you can do:
switch(c) {
case CONST_STAR:
...
write(fd, &c_star, sizeof(c_star))
...
}
(Note also that sizeof(char) is always one, by definition. You may know that already, but it's not as widely appreciated as perhaps it should be.)
Here is another way to solve this old but still relevant question
#define THIS_CONST 'A'
//I place this in a header file
static inline size_t write1(int fd, char byte)
{
return write(fd, &byte, 1);
}
//sample usage
int main(int argc, char * argv[])
{
char s[] = "Hello World!\r\n";
write(0, s, sizeof(s));
write1(0, THIS_CONST);
return 0;
}
Ok, I've come up with a bit of a hack, for chars only - but I'll put it here to see if it inspires any better solutions from anyone else
static const char *charptr(char c) {
static char val[UCHAR_MAX + 1];
val[(unsigned char)c] = c;
return &val[(unsigned char)c];
}
...
write(fd, charptr(THIS_CONST), sizeof(char));

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