So I'm learning to code (in C) for a few months now and I'm focusing here on memory allocation.

My code is very simple, everything works, and yet at the end, it doesn't return a 0 but instead I get a -1073740940 (0xC0000374) which, I noticed, have something to do with memory allocation (lol). And seriously I don't know how to correct it.

Here it is :

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

int main()
{

 char* chaine1 = (char*)malloc(10*sizeof(char));
 char* chaine2 = (char*)malloc(10*sizeof(char));

 if(chaine1 == NULL || chaine2 == NULL)
    return -2;

 strcpy(chaine1, "Salut les");
 strcpy(chaine2, " codeurs!");

 printf("Chaine 1: %s\n", chaine1);
 printf("Chaine 2: %s\n", chaine2);

 int taille1 = strlen(chaine1);
 int taille2 = strlen(chaine2);

 char* tmp_chaine = (char*)malloc(sizeof(char)*10);

 if(tmp_chaine == NULL)
    return -2;

 strcpy(tmp_chaine, chaine1);

 realloc(chaine1, sizeof(char)*(taille1+taille2+1));

 for(int i = 0; i <= taille1; i++)
    chaine1[i] = tmp_chaine[i];

 for(int i = 0; i <= taille2; i++)
    chaine1[taille1+i] = chaine2[i];

printf("%s", chaine1);

return 0;
}

I noticed that when I try to free my strings, the program crashes before the end, but that's all I could figure out.

(I know the code has nothing optimized, please don't yell at me :( )

So I'm learning to code (in C) for a few months now and I'm focusing here on memory allocation.

My code is very simple, everything works, and yet at the end, it doesn't return a 0 but instead I get a -1073740940 (0xC0000374) which, I noticed, have something to do with memory allocation (lol). And seriously I don't know how to correct it.

Here it is :

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

int main()
{

 char* chaine1 = (char*)malloc(10*sizeof(char));
 char* chaine2 = (char*)malloc(10*sizeof(char));

 if(chaine1 == NULL || chaine2 == NULL)
    return -2;

 strcpy(chaine1, "Salut les");
 strcpy(chaine2, " codeurs!");

 printf("Chaine 1: %s\n", chaine1);
 printf("Chaine 2: %s\n", chaine2);

 int taille1 = strlen(chaine1);
 int taille2 = strlen(chaine2);

 char* tmp_chaine = (char*)malloc(sizeof(char)*10);

 if(tmp_chaine == NULL)
    return -2;

 strcpy(tmp_chaine, chaine1);

 realloc(chaine1, sizeof(char)*(taille1+taille2+1));

 for(int i = 0; i <= taille1; i++)
    chaine1[i] = tmp_chaine[i];

 for(int i = 0; i <= taille2; i++)
    chaine1[taille1+i] = chaine2[i];

printf("%s", chaine1);

return 0;
}

I noticed that when I try to free my strings, the program crashes before the end, but that's all I could figure out.

(I know the code has nothing optimized, please don't yell at me :( )

• c
• memory-management
• variable-assignment

• In realloc(chaine1, sizeof(char)*(taille1+taille2+1)); you've thrown away the new pointer, and the old pointer chaine1 is now dead. But you continue to use it. – Weather Vane Commented Feb 1 at 17:51
• when i try to free my strings isn't shown, but you can't free(chaine1) because it has already been freed by realloc. – Weather Vane Commented Feb 1 at 17:55
• look at using valgrind and cppcheck tools to assist you whilst developing code – ticktalk Commented Feb 1 at 23:29

2 Answers 2

Code 0xC0000374 is STATUS_HEAP_CORRUPTION. The heap corruption happens because realloc() is used improperly.

realloc(chaine1, sizeof(char)*(taille1+taille2+1));

The returned new pointer is ignored, chaine1 pointer is unchanged and points to an old and freed memory location. Further lines use that wrong memory location and corrupt the heap.

You can allocate the required size a few lines above with malloc() without using realloc().

If you still wish to try realloc() then it should be

char* new_chaine1 = realloc(chaine1, taille1+taille2+1);
if (new_chaine1) {
  chaine1 = new_chaine1;
  new_chaine1 = NULL;
} else {
  // handle the reallocation error
}

The code exhibits memory leaks in main() without free().

Further notes. sizeof(char) is odd, always 1, can be omitted. (char*) cast from void* is not needed at malloc().

hmmmmm

$ make so
cc -O2 -pipe  so.c  -o so
so.c:30:2: warning: ignoring return value of function declared with 'warn_unused_result' attribute [-Wunused-result]
   30 |         realloc(chaine1, sizeof(char)*(taille1+taille2+1));
      |         ^~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1 warning generated.

This warning, is indeed a serious error, as if realloc() needs to reallocate the memory in another place, it will return a valid pointer pointing to a different place, invalidating the one you passed to the routine.

You can do:

chaine1 = realloc(chaine1, sizeof(char)*(taille1 + taille2 + 1));

instead. If you don't and realloc() moves the buffer to elsewhere, the memory pointed by chaine1 will be unallocated, and Undefined Behavior will be in force. Look that I check the pointer again, after realloc() has return, as it could be wrong (the original value will have been lost if you don't save it first, but you cannot continue, once realloc has failed)

Next is how your code should appear, after the modifications proposed:

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

int main()
{   
    char* chaine1 = (char*)malloc(10*sizeof(char));
    char* chaine2 = (char*)malloc(10*sizeof(char));
   
    if(chaine1 == NULL || chaine2 == NULL)
        return -2;
   
    strcpy(chaine1, "Salut les");
    strcpy(chaine2, " codeurs!");
   
    printf("Chaine 1: %s\n", chaine1);
    printf("Chaine 2: %s\n", chaine2);
   
    int taille1 = strlen(chaine1);
    int taille2 = strlen(chaine2);
   
    char* tmp_chaine = (char*)malloc(sizeof(char)*10);
   
    if(tmp_chaine == NULL)
        return -2;
   
    strcpy(tmp_chaine, chaine1);
   
    chaine1 = realloc(chaine1, sizeof(char)*(taille1+taille2+1));

    if (chaine1 == NULL)
        return -3;
   
    for(int i = 0; i <= taille1; i++)
        chaine1[i] = tmp_chaine[i];

    for(int i = 0; i <= taille2; i++)
        chaine1[taille1+i] = chaine2[i];

    printf("%s", chaine1);

    return 0;
}