CODEWITHSUNDEEP
cpointers64-bitmemory-address
June Yoon
June Yoon

Reputation: 512

I don't understand why some pointer addresses are like this

I'm studying pointer in C language and I have some questions.

#include <stdio.h>

int main()
{
    char c = 'A';
    char* pc = &c;
    char** ppc = &pc;

    printf("%p %p\n", pc, ppc);
    printf("%p %p\n", pc + 1, ppc + 1);
    printf("%p %p\n", &c, &c + 1);
    printf("%p %p\n", &pc, &ppc);
    printf("%p %p\n", &pc + 1, &ppc + 1);

    return 0;
}

In this code, let's say that

&c = 0117FE7B
&pc = 0117FE6C
&ppc = 0117FE60

I thought some answers will be like this:

ppc + 1 = 0117FE6D
&pc + 1 = 0117FE6D
&ppc + 1 = 0117FE61

but the correct answer was like this:

ppc + 1 = 0117FE70
&pc + 1 = 0117FE70
&ppc + 1 = 0117FE64

and I don't understand why. Can someone explain this for me? (My computer is using 64 bit windows OS.)

Upvotes: 1

Views: 91

Answers (1)

John Bode
John Bode

Reputation: 123468

Pointer arithmetic is done in terms objects, not bytes. If p evaluates to the address of a 4-byte int object, then p + 1 evaluates to the address of the next 4-byte int, not the next byte:

int x;           // assume 4-byte int
int *ip = &x;

short s;         // assume 2-byte short
short *sp = &s;

char c;
char *cp = &c;

    +---+                  +---+                   +---+
x : |   | <-- ip       s : |   | <-- sp        c : |   | <-- cp
    +---+                  +---+                   +---+
    |   |                  |   |                   |   | <-- cp + 1
    +---+                  +---+                   +---+
    |   |                  |   | <-- sp + 1        |   |
    +---+                  +---+                   +---+
    |   |                  |   |                   |   |
    +---+                  +---+                   +---+
    |   | <-- ip + 1       |   |                   |   |
    +---+                  +---+                   +---+
    |   |                  |   |                   |   |
    +---+                  +---+                   +---+
    |   |                  |   |                   |   |
    +---+                  +---+                   +---+
    |   |                  |   |                   |   |
    +---+                  +---+                   +---+

So depending on the size of the pointed-to type, p + 1 will either give the address + 1, or the address + 4, or the address + 8, etc.

Remember, the array subscript operation a[i] is defined as *(a + i) - given a starting address a, offset i objects (not bytes!!) from that address and deference the result.

Upvotes: 5

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