Why the realloc did not work properly in this case?

Question

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

char *func(char * str){


    int len;
    len=strlen(str)+3;
    str = (char *)realloc(str,len);
    return str;


}

void main(){

    printf("str:%s",func("hello"));

}

The final ans prints (null),instead of printing the string: "hello". Can anyone please explain why is it so? I am unable to identify any error. Can anyone rectify the error, and help me with a working code. Please!

You call realloc on a char * that can't be realloced. So it fails, and returns NULL. — William Pursell, Jul 04 '16 at 15:05
The question is tagged `dynamic-allocation`, and yet there's none in sight... @WilliamPursell Shouldn't this generate a warning, or complete failure to compile, for passing a `char const *` (type of string literal) to `func()`? I mostly program C++14 with all warnings on, so maybe I'm too used to good type safety... — underscore_d, Jul 04 '16 at 15:06
@2501 Then I guess that's a C thing. More's the pity. In C++, implicitly converting `char const *` to `char *` is still legal but deprecated, and many compilers will warn the programmer into oblivion for trying it: http://stackoverflow.com/a/2760547/2757035 And rightly so! Any attempt to edit that [scare quotes] '`char *`' is UB, and too easy to try. — underscore_d, Jul 04 '16 at 15:13
__Can anyone rectify the error, and help me with a working code. Please!__ Sorry, did you actually take ypour time to read the given answers? We're not here to write the code for you, we can only show you the way... — Sourav Ghosh, Jul 04 '16 at 15:13
@underscore_d I don't get your point. You can't convert const to non-const in C without a cast either. — 2501, Jul 04 '16 at 15:15
@2501 My point is that C++ is stricter and defines string literals as `char const *`, whereas C just uses `char *`. C++ is seemingly also dropping the ability to silently convert and drop the const. — underscore_d, Jul 04 '16 at 15:23
@underscore_d `C++ is seemingly also dropping the ability to silently convert and drop the const` There is no such feature in C. You can't convert const to non-const in C without a cast. — 2501, Jul 04 '16 at 15:25
@2501 Please read again and let me know if I've misunderstood. In C, string literals are not `const`. In C++, they are `const` _but_ could be implicitly converted in contexts such as this. In C++, this is now strongly deprecated or even forbidden, depending on which compiler you ask. — underscore_d, Jul 04 '16 at 15:28
[Don't cast the result of `malloc` in C](http://stackoverflow.com/q/605845/995714) — phuclv, Jul 04 '16 at 15:32

score 6 · Accepted Answer · edited May 23 '17 at 10:28

Your program invokes undefined behavior because you're passing a pointer, which was not previously returned by dynamic memory allocator family of functions, to realloc().

According to C11, chapter §7.22.3.5, The realloc function, (emphasis mine)

If ptr is a null pointer, the realloc function behaves like the malloc function for the specified size. Otherwise, if ptr does not match a pointer earlier returned by a memory management function, or if the space has been deallocated by a call to the free or realloc function, the behavior is undefined. [...]

That said,

For a hosted environment, void main() should better be int main(void), at least.
Please see this discussion on why not to cast the return value of malloc() and family in C..

@Anonymous [prefered way at stackoverflow to say "thanks"](http://meta.stackexchange.com/a/5235/202811) — Grijesh Chauhan, Jul 04 '16 at 15:46

score 6 · Answer 2 · answered Jul 04 '16 at 15:08

"hello" is a read-only string literal. Its type is really a const char* although compilers allow assignment to a char* despite all the merry hell that this causes.

The behaviour on calling realloc on such a pointer is undefined. So the compiler can do anything it likes. It might eat your cat.

Vlad from Moscow · Answer 3 · 2016-07-04T15:16:41.143

1

You may realloc an object that was allocated dynamically. String literals have static storage duration and may not be changed. Any attempt to modify a string literal results in undefined behavior.

What you could do is the following

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *func( const char * str )
{
    size_t len = strlen( str ) + 3;

    char *tmp = realloc( NULL, len );

    if ( tmp ) strcpy( tmp, str );

    return tmp;
}

int main( void )
{
    char *str = func( "hello" );

    if ( str ) printf( "str: %s\n", str );

    free( str );
}

The program output is

str: hello

Take into account that the call

realloc( NULL, len )

is equivalent to

malloc( len )

edited Jul 04 '16 at 15:16

answered Jul 04 '16 at 15:06

Vlad from Moscow

301,070
26
186
335

`realloc ( NULL, len );` Why not just `malloc(len)`? They're equivalent, but the latter seems better semantically. – underscore_d Jul 04 '16 at 15:15
@underscore_d Because the question about realloc. – Vlad from Moscow Jul 04 '16 at 15:16

Why the realloc did not work properly in this case?

3 Answers3