Question

when i run this i get segv on printf, what am i doing wrong?

int main() {
        char **bla;
        int size =10;
        int i;

        bla = calloc(size*size,sizeof(char *));

        for(i=0;i<size;i++) {
                *bla = calloc(10,sizeof(char));
                strncpy(*bla,"aaaaa",size);
                bla++;
        }

        printf("%s\n",bla[0]);
}

I know i could do this with

int main() {
        char **bla;
        int size =10;
        int i;

        bla = calloc(size*size,sizeof(char *));

        for(i=0;i<size;i++) {
                bla[i] = calloc(10,sizeof(char));
                strncpy(bla[i],"aaaaa",size);
        }



   printf("%s\n",bla[0]);
   }

but is there any way to do this with pointers?

Answer 1

By writing bla++, you’re changing bla to point to the next pointer.

At the end, bla[0] (which is equivalent to *bla) has been incremented 10 times and will point the memory location immediately after the allocated block.

You could fix this by writing bla -= 10 after the loop.

However, the best way to fix it is to not increment bla at all, and instead write

 *(bla + i) = calloc(10,sizeof(char));

Alternatively, you could declare a second pointer (char** currentBlah = blah) and increment it instead, then print blah[0], which will still point to the original memory location.