Do we lose data in a buffer after realloc'ing?

Question

I'm having troubles understanding how realloc works. If I malloc'ed a buffer and copied data to that buffer, let's say "AB":

 +------------+
 | A | B | \0 |
 +------------+

then I realloc'ed the buffer, will there be any lost in the data (even a single byte)?; or it just does expanding the buffer? :

 +------------------------+
 | A | B | \0 | ? | ? | ? |
 +------------------------+

code:

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

int main(void){

    char* buffer    = (char*) malloc( sizeof(char) * 3 );
    strncpy(buffer, "AB", 2);

    buffer          = (char*) realloc(buffer, sizeof(char) * 6); /* Will there be any lost here? */
    free(buffer);
    return(0);
}

David Heffernan · Accepted Answer · 2012-02-04T17:37:16.550

68

A realloc that increases the size of the block will retain the contents of the original memory block. Even if the memory block cannot be resized in placed, then the old data will be copied to the new block. For a realloc that reduces the size of the block, the old data will be truncated.

Note that your call to realloc will mean you lose your data if, for some reason the realloc fails. This is because realloc fails by returning NULL, but in that case the original block of memory is still valid but you can't access it any more since you have overwritten the pointer will the NULL.

The standard pattern is:

newbuffer = realloc(buffer, newsize);
if (newbuffer == NULL)
{
    //handle error
    return ...
}
buffer = newbuffer;

Note also that the casting the return value from malloc is unnecessary in C and that sizeof(char) is, by definition, equal to 1.

edited Feb 04 '12 at 17:37

answered Feb 04 '12 at 17:25

David Heffernan

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1

Why can't you do `buffer = realloc(buffer, newsize);` ? – ylun.ca Oct 01 '15 at 17:03
2

@ylun for the reasons explained in the answer – David Heffernan Oct 01 '15 at 18:25
Right so in order to prevent data loss you would first check that the allocation is successful and then reassign, thanks. – ylun.ca Oct 01 '15 at 18:35
1

What if you realloc a smaller block of memory? – Kenny Worden Apr 30 '17 at 21:00
Obviously you lose what you deallocate but you keep the rest – David Heffernan Apr 30 '17 at 21:01

pmg · Answer 2 · 2012-02-04T17:36:00.223

Nothing is lost. But you really should test if the realloc() (and the malloc() before) "worked".
Also the cast to the return value of malloc is, at best, redundant, and it may hide an error the compiler would have caught in its absence.

based on the assumption you want strings, your usage of strncpy is wrong

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

int main(void) {
    char *buffer = malloc(3);
    if (buffer == NULL) /* no memory */ exit(EXIT_FAILURE);

    strncpy(buffer, "AB", 2);
    /* ATTENTTION! ATTENTION: your array is not a string.
    ** buffer[2] is not the zero string terminator */

    // buffer = realloc(buffer, 6); /* Will there be any lost here? */
    /* If realloc returns NULL, you've just lost the only pointer to
    ** the allocalted memory, by overwriting it with NULL.
    ** Always `realloc` to a temporary variable */
    char *tmp_buffer = realloc(buffer, 6);
    if (tmp_buffer == NULL) {
        /* realloc failed */
    } else {
        /* realloc worked, no bytes lost */
        buffer = tmp_buffer;
        /* ATTENTION! ATTENTION: buffer is still not a string
        ** buffer[0] is 'A', buffer[1] is 'B',
        ** all other elements of buffer are indeterminate */
    }

    free(buffer);
    return(0);
}

Do we lose data in a buffer after realloc'ing?

2 Answers2

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