Pointer to an integer array

Question

Suppose mat is a pointer to an array of size 5 where each element is an integer

int (*mat)[5];

and I have initialized it as

int a[5] = {5, 4, 3, 2, 1};
mat = &a;

I've written the code as

#include <stdio.h>

int main()
{
    int (*mat)[5];
    int a[5] = {5, 4, 3, 2, 1};
    mat = &a;
    printf("%p\n%p\n%d\n", mat, *mat, **mat);
    return 0;
}

Output:

43800  
43800  
5

Why does mat and *mat give the same answer?

@JGroven Array decaying means that, when an array is passed as a parameter to a function, it's treated identically to ("decays to") a pointer. However, there is no any passing case ?? — Soner from The Ottoman Empire
– Soner from The Ottoman Empire, Commented Aug 8, 2018 at 17:11
@snr I was under the impression that array decaying is the implicit conversion from array to pointer, and that using an array as a parameter is merely a demonstration of decay. — JGroven
– JGroven, Commented Aug 8, 2018 at 17:19
@JGroven ((: just was kidding, Thank you for explanation freshman — Soner from The Ottoman Empire
– Soner from The Ottoman Empire, Commented Aug 8, 2018 at 17:22
Arrays decay to pointers in nearly every case. The two exceptions are the sizeof and the & operator. — Osiris
– Osiris, Commented Aug 8, 2018 at 17:23

John Bode · Accepted Answer · 2018-08-08 19:18:33Z

A picture may help:

     +---+            +---+
mat: |   | ------> a: | 5 | a[0]
     +---+            +---+
                      | 4 | a[1]
                      +---+
                       ...
                      +---+
                      | 1 | a[4]
                      +---+

So, first thing we notice - the address of the array a is the same as the address of the array element a[0]. The expression a has type "5-element array of int"; unless that expression is the operand of the sizeof or unary & operators, it is converted ("decays") to an expression of type "pointer to int", and the value of the expression is the address of the first element of the array. Thus, the expressions &a, a, and &a[0] will all yield the same value (address of the first element of a), even though they're not all the same type (int (*)[5], int *, int *).

So, given the above, all of the following are true:

  mat == &a 
 *mat ==  a
**mat ==  a[0]

Since &a and a evaluate to the same address, mat and *mat evaluate to the same value.

Quentin · Accepted Answer · 2018-08-09 08:43:11Z

1

Quite simply, the address of an array is the same as its first element's. In this context, *mat is of type int[5] which decays to int *, i.e. a pointer to the first element of the array.

edited Aug 9, 2018 at 8:43

Quentin

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answered Aug 8, 2018 at 17:07

FatalError

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3 Comments

metasj Over a year ago

if I have an array int a[5], a is an integer pointer containing &a[0],but what is the address of a?

Osiris Over a year ago

@metasj &a and &a[0] have a different type. a is of type int[5]. &a[0] is of type int *. &a is of type int (*)[10]. Arrays are not pointers.

Ian Abbott Over a year ago

@metasj a is not an integer pointer but it will evaluate as an integer pointer in most kinds of expressions.

chux · Accepted Answer · 2018-08-09 01:53:34Z

mat is a pointer to array 5 of int.

*mat is an array 5 of int.

Arrays such as *mat, when pass to functions are converted in "... to an expression with type pointer to type that points to the initial element of the array object...". C11 §6.3.2.1 3

*mat in printf("%p\n",*mat); is a pointer to an int.

"Why does mat and *mat give the same answer?"

These both point to the same memory location. mat points to the array. In printf(), *mat points to the first array element, an int.

As pointers, mat and *mat point to very different types. They could have employed different presentation (even size) and yielded different text output, yet their values are equivalent. This difference is uncommon.

Further, "%p" expects a void*. Use a void* cast for portable code.

printf("%p\n%p\n",mat,*mat);  // undefined behavior
printf("%p\n%p\n",(void*) mat,(void*) *mat);

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