Using pointer with array

Question

I am learning pointers in C programming. A piece of code is:

#include <stdio.h>
    int main()
    {
        int ary[4] = {1, 2, 3, 4};
        int *p = ary + 3;
        printf("%d\n", p[-2]);
    }

ary is the array and how is the *p pointing to the ary+3 and output coming is 2. How is the pointer handling the array values? I looked into https://www.geeksforgeeks.org/pointer-array-array-pointer/ but I did not found how pointer is influencing array there?

The pointer arithmetic works in a similar way to array indexing. The array decays to a pointer, so ary + 3 has the same value as &ary[3]. — Weather Vane
– Weather Vane, Commented Aug 8, 2021 at 16:11
i tried to see the different result by changing index. p[0] gives '4' and p[1],p[2] gives '0' how is this seen? — AB21
– AB21, Commented Aug 8, 2021 at 16:17
p[1] and p[2] are indexing beyond the last element of the original array, which is undefined behaviour and the values found are irrelevant. — Weather Vane
– Weather Vane, Commented Aug 8, 2021 at 16:19
https://www.geeksforgeeks.org/pointer-array-array-pointer/ has some really bad code posted. sizeof(p) = %lu is undefined behavior and flat out wrong on 64-bit Windows systems where size_t is 64 bits and long is 32 bits. — Andrew Henle
– Andrew Henle, Commented Aug 8, 2021 at 16:24

Shreekesh Murkar · Accepted Answer · 2021-08-08 16:19:44Z

4

It is very easy to understand.

We know arris an array but the arr reference will always point to an address of a starting element of the array.

So, for above example. arr is point to address of arr[0] (value 1). arr + 3 will point to the arr[3] (value 4).

That means pointer p is point to arr[3]. Now, p[-2] will move the pointer 2 address back and will point to arr[1] (value 2).

Hence the output 2 !

edited Aug 8, 2021 at 16:19

answered Aug 8, 2021 at 16:17

Shreekesh Murkar

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alex01011 · Accepted Answer · 2021-08-08 16:28:38Z

2

The array syntax p[-2] is equivalent to, *(p - 2), since p is pointing to ary + 3 which is the fourth element in your array, decrementing the pointer by 2, will print 2.

p[0] will print the value p is pointing to, since *(p + 0), any other positive index will result in accessing the array out of bounds, which is not defined.

edited Aug 8, 2021 at 16:28

answered Aug 8, 2021 at 16:19

alex01011

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John Bode · Accepted Answer · 2021-08-08 18:11:34Z

The subscript expression a[i] is defined as *(a + i) - given an address a¹, offset i elements (not bytes! - more on that below) from that address and dereference the result. ary[0] is equivalent to the expression *(ary + 0), which is equivalent to *ary, ary[1] is equivalent to *(ary + 1), etc.

The expression p[-2] is equivalent to *(p - 2); since p == ary + 3, then p - 2 == ary + 3 - 2 == ary + 1, which is the address of the second element of ary. Thus, p[-2] == ary[1] == 2.

Pointer arithmetic is based on the size of the pointed-to type - if p contains the address of an object of type T, then p + 1 yields the address of the next object of that type. For example, given

char  *cp = (char *)  0x1000;
short *sp = (short *) 0x1000;
long  *lp = (long *)  0x1000;

then we have the following:

Address    char            short          long
-------    +---+           +---+          +---+         
0x1000     |   | cp        |   | sp       |   | lp
           +---+           + - +          + - +
0x1001     |   | cp + 1    |   |          |   |
           +---+           +---+          + - +
0x1002     |   | cp + 2    |   | sp + 1   |   |
           +---+           + - +          + - +
0x1003     |   | cp + 3    |   |          |   |
           +---+           +---+          +---+
0x1004     |   | cp + 4    |   | sp + 2   |   | lp + 1
           +---+           + - +          + - +
            ...             ...            ...

^{Arrays are not pointers - however, unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted, or "decay", to an expression of type "pointer to T" and the value of the expression will be the address of the first element of the array.}

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