(C-Program) example Code about Array Address

Question

#include <stdio.h>

#include <sdint.h>

int32_t a = 0;

void do_stuff(int32_t* c){
    int32_t static b = 2;
    printf("Address: %p\n", c+a);
    printf("%d %d\n", *(c+a),c[b]);
    a+=4;
    b+=b^b;
}

int main() {
    int32_t array[9] = {42, 5, 23, 82, 127, 21, 324, 3, 8};
    printf("Array Address: %p\n", array);
    do_stuff(array);
    do_stuff(array);
    return =0;
}

^ means XOR Operation
assume that the printf on line printf("Array Address: %p\n", array); gives you this: Array Address: 0x08

Question: What is the output of this program?

Answer:

0x08

42, 23

0x18

127, 23

My question : why is the output of printf("Address: %p\n", c+a); for the second do_stuff is 0x18?

Note that `b ^ b` is zero. And will always be. – Adrian Mole Jul 18 '22 at 12:03 — Adrian Mole, Jul 18 '22 at 12:03

score 1 · Accepted Answer · answered Jul 18 '22 at 12:11

1

If not take into account typos in the presented code then after this statement

a+=4;

the expression c + a points to the fifth element of the array that is the offset from the beginning of the array is equal to 4 * sizeof( int32_t ) that is the same as 4 * 4 and in hex is equal to 0x10.

So 0x10 + 0x8 == 0x18

answered Jul 18 '22 at 12:11

Vlad from Moscow

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why do we turn 4*4 to hex ? – student Jul 20 '22 at 19:39
@student If to use the function printf to output an address then the conversion specifier %p is used that outputs addresses in hex. – Vlad from Moscow Jul 20 '22 at 19:40

(C-Program) example Code about Array Address

1 Answers1