Incrementing pointers in C arrays

Question

I'm trying to understand the below program. Specifically, the definition of cur_name and the incrementing of the pointer for cur_age in the printf statement.

*(cur_age + i) must be indexing each of the integers in the array but I would have expected this to point to successive addresses in memory and not the next integer given ints are 4 bytes? i.e. why not i+4

#include <stdio.h>

int main(int argc, char *argv[])
{
    int ages[] = {23, 43, 12, 89};
    char *names[] = {"Anne", "Kay", "Joe", "Pete"};
    int count  = sizeof(ages) / sizeof(int);
    int *cur_age = ages;
    char **cur_name = names;

    for (int i = 0; i < count; i++) {
        printf("%s is %d years old.\n", *(cur_name + i), *(cur_age + i));
    }

    return 0;
}

SergeyA · Accepted Answer · 2016-05-20 16:34:37Z

6

This is simply how the pointer arithmetics is defined in C. Incrementing the pointer always considered to be incrementation in units, where one unit is the sizeof() of type being pointed to.

Side note - while int's are commonly 4 bytes in size, this is not set in stone. They might as well be smaller (2 bytes) or longer (really not limit).

answered May 20, 2016 at 16:34

SergeyA

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3 Comments

CatsLoveJazz Over a year ago

So when incrementing a pointer it takes into account the type, great! As for the character array, I understand this is a pointer to an array of pointers but why the **.

SergeyA Over a year ago

@CatsLoveJazz, not really. It is an array of pointers (not a pointer to such array). Every element of this array is a pointer to char*, with size of char* size (for example, 8 bytes). Arrays can decay to the pointers (in fact, they do it almost always), so when you say char** cur_name = names names is decayed into pointer to pointer to char.

CatsLoveJazz Over a year ago

I will have to digest that, thanks. I was referring to the cur_name variable.

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