Question

Getting double free for below code, if a long string passed.
I tried all sorts of things. If I remove the free(s) line it goes away.
Not sure why it is happening.

void format_str(char *str1,int l,int o) {
    char *s = malloc(strlen(str1)+1);
    char  *s1=s, *b = str1;
    int i=0;
    while(*str1!='\0') {
        i++;
        *s1++=*str1++;

        if(i>=l) {
            if(*str1!=',') {
                continue;
            }
        *s1++=*str1++;
            *s1++='\n';
            for(i=0;i<o;i++) {
                *s1++=' ';
            }
            i = 0;
        }
    }
    *s1 = '\0';
    strcpy(b,s);
    free(s);
}

Answer 1

You are almost certainly corrupting the heap. For example:

int main() 
{
    char original[1000] = "some,,,string,,, to,,,,format,,,,,";

    printf( "original starts out %u characters long\n", strlen(original));
    format_str( original, 6, 6);
    printf( "original is now %u characters long\n", strlen(original));

    return 0;
}

would require that the buffer allocated by malloc() be much larger than strlen(str1)+1 in size. Specifically, it would have to be at least 63 bytes long (as the function is coded in the question, the allocation has a size of 35 bytes).

If you need more specific help, you should describe what you’re trying to do (such as what are the parameters l and o for?).