C：SEGFAULT与char **上的realloc

Question

Another one in my series of problems with this code. I have below function which is comparing arg with every string in the array of strings reference :

char compare(char *arg)
{
        int iter=0;
        char retchar='0';

        while(iter < no_of_ref)
        {
        //      printf("arg : %s , reference : %s \n",arg,reference[iter]);
                if((strstr(reference[iter],arg) != NULL) || (strstr(arg,reference[iter]) != NULL))
                {
                        retchar='1';
                        break;
                }
          iter++;
        }
return retchar;
}

reference is global char ** , but built up dynamically inside main as below:

reference = calloc(CHUNK, sizeof(char *));

Then some code, then:

                        temp_in[pre_pip+1]='\0';
                        reference[no_of_ref]=malloc(strlen(temp_in) + 1);
                        strcpy(reference[no_of_ref++],temp_in);
                        memset(&temp_in,'\0',sizeof(temp_in));
                        pre_pip = -1;
   printf("INDEX: %d, address : %p , val : %s\n",no_of_ref-1,reference[no_of_ref-1],reference[no_of_ref-1]);      //DEBUG
                }
                /*If allocated buffer is at brim, extend it for CHUNK char *  further*/
                if(no_of_ref == (tr*CHUNK - 2))
                {
                        current_size = tr*CHUNK*sizeof(char *);

                        char *retalloc = realloc(reference,current_size + CHUNK*sizeof(char *));
                                if(retalloc == NULL)
                                        perror("ERROR on realloc");
                                else
                                {
                                        printf("Realloced successfully: %p\n",retalloc);
                                        tr++;
                                }

The code running fine for test case where no need to realloc arises, ie Number of input strings is less than CHUNK . In case of realloc , I'm getting SEGFAULT from function. Below is for one of the run:

Program terminated with signal 11, Segmentation fault.
#0  __strstr_sse42 (s1=0x3839393433333230 <Address 0x3839393433333230 out of bounds>, s2=0x6020c0 <cmp> "8956549122") 

Answer 1

You need to put parenthesis for expression in realloc() as

//---------------------------------v -------------------v
char *retalloc = realloc(reference,(current_size + CHUNK)*sizeof(char *));

Assume CHUNK=100 and current_size=200 , your code will allocate 200+100*8=1000 bytes while you want (200+100)*8 = 2400 bytes

Also, make sure you assign retalloc to reference variable after reallocation.

Answer 2

When realloc actually reallocates the memory you pass to it, then that pointer you pass as an argument still points to the old memory area. The realloc function returns a pointer to the new memory, so you have to assign that to eg reference .