referenced array == array

Question

could you tell my why the value of a referenced array and the value of the array itself has the same value?

i know a is a type of int* but with &a it should be int** or am i wrong?? so the value should be a pointer to the a int pointer. example code:

    #include &ltstdio.h>

    int main()
    {
        int a[10];
        printf("a without ref.: 0x%x\n",a);
        printf("a with    ref.: 0x%x\n",&a);
        return 0;
    }

http://ideone.com/KClQJ

possible duplicate of [C: How come an array's address is equal to its value?](http://stackoverflow.com/questions/2528318/c-how-come-an-arrays-address-is-equal-to-its-value) — jpalecek, Feb 16 '12 at 13:11
You might like [an answer of mine to a similar question](http://stackoverflow.com/questions/5181237/confusion-in-data-types-in-a-2d-array/5181495#5181495) asked some time ago. — pmg, Feb 16 '12 at 13:20

score 3 · Accepted Answer · answered Feb 16 '12 at 13:09

3

Name of the array decays to an pointer to its first element in this case.

answered Feb 16 '12 at 13:09

Alok Save

202,538
53
430
533

I don't think it's the *name* that decays. The array itself decays :) +1 anyway – Armen Tsirunyan Feb 16 '12 at 13:41

score 1 · Answer 2 · answered Feb 16 '12 at 14:03

Name of the array will implicit convert to a pointer ,except for two situation ,the one situations is "&array",the other is "sizeof(array)".In both cases,name of the array is a array ,not a pointer .

For example:

int a[10];
int *p;
p = a;        //a is a pointer
p = &a;       //a is a array,&a is a constant pointer
sizeof(a);    //a is array

score 1 · Answer 3 · answered Feb 16 '12 at 14:26

Given an array declaration T a[N], the expression &a has type "pointer to N-element array of T (T (*)[N]) and its value is the base address of the array. In that respect, the unary & operator behaves the same for arrays as it does for any other data type.

What's hinky is how C treats the array expression a. Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an array expression of type "N-element array of T" (T [N]) will be replaced with ("decay to") a pointer expression of type "pointer to T" (T *) and its value will be the address of the first element of the array. IOW, a == &a[0].

Since the address of the first element of the array is the same as the base address of the entire array, the expressions a and &a yield the same value, but the types are different (T * as opposed to T (*)[N]).

score 0 · Answer 4 · answered Feb 16 '12 at 13:28

0

if there is int *p which point a, then,

a+0 == &a[0] == p+0 == &p[0] : address

*(a+0) == *(&a[0]) == *(p+0) == *(&p[0]) : data

a == &a == &a[0]

answered Feb 16 '12 at 13:28

plan9assembler

2,862
1
24
13

The **types** of `a` and `&a` are not the same (though the values are): you cannot say that 42 carrots is the same as 42 potatoes. – pmg Feb 16 '12 at 14:40

referenced array == array

4 Answers4