Explain 2-D array using pointer in c

Question

I am trying to run the following code and getting "gh" as the output. Please help me out, how are we getting this result.

void f(char **p) {
    char *t;
    t = (p += sizeof(int))[-1];
    printf("%s\n", t);
}

int main() { 
    char *arg[] = { "ab", "cd", "ef", "gh", "ij", "kl" };
    f(arg);
}

P.S. I am using the gcc compiler.

haccks · Accepted Answer · 2017-02-12 19:59:37Z

1

p+=sizeof(int) will make p to point sizeof(int) elements farther from the 0^th element so that p+=sizeof(int) will give the address of the 4^th element of passed array (assuming size of int 4 bytes on your machine). [-1] will decrement it by one letting t to point to the 3^rd element which is "gh".

edited Feb 12, 2017 at 19:59

answered Feb 12, 2017 at 19:37

haccks

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2 Comments

chqrlie Over a year ago

p+=sizeof(int) will add sizeof(int) bytes to p: Not really, it will make p point sizeof(int) elements farther. p is a char **, so the elements are char*. The rest of your explanation is correct.

haccks Over a year ago

@chqrlie; I need some rest for today.

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