why in array a[] , &a and a are the same?

Question

In this program:

#include <stdio.h>

int main()
{
    int a[2];
    int *b=&a;
    printf("%p\n",a);
    printf("%p\n",&a);
    printf("%p",b);
    return 0;
}

All printing same things!! a is a pointer to first place of our array right?! It must be saved somewhere!! so &a or b must be a pointer to a...

I'm wrong?!

BTW ive got a warning from my compiler about int *b=&a;

[Warning] initialization from incompatible pointer type [enabled by default]

Read [Difference between `&str` and `str`, when `str` is declared as `char str[10]`?](http://stackoverflow.com/questions/15177420/what-does-sizeofarray-return/15177499#15177499) — Grijesh Chauhan, Jan 13 '14 at 08:36
And I think I have answer your question here [Inconsistency in using pointer to an array and address of an array directly](http://stackoverflow.com/questions/17661078/inconsistency-in-using-pointer-to-an-array-and-address-of-an-array-directly/17663091#17663091) It is kind of duplicate. — Grijesh Chauhan, Jan 13 '14 at 08:38
@anishsane he didn't miss `[0]` but OP thinks that `&a` is address of first element...Read title of question he is asking `a` and `&a` are same?? — Grijesh Chauhan, Jan 13 '14 at 11:18
I meant, typically, people ask differences between `a`, `&a` & `&a[0]`... OP asked only 2... — anishsane, Jan 13 '14 at 11:47

Jeyaram · Accepted Answer · 2014-01-13T08:46:10.470

0

&a is int (*)[2]. i.e pointer to an array.

But int *b=&a; assigns to integer pointer. Though it points to proper address, you need to typecast to remove the warning message.

Add (int*) like.

int *b=(int*)&a;

#include <stdio.h>

int main()
{
    int a[2] = {10,20};
    int *b=(int*)&a;

    printf("\n :%d   %d\n", b[0], b[1]);
#if 0
    printf("%p\n",a);
    printf("%p\n",&a);
    printf("%p",b);
#endif
    return 0;
}

prints 10 and 20.

edited Jan 13 '14 at 08:46

answered Jan 13 '14 at 08:32

Jeyaram

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1

Your cast is incorrect. To get valid code you must either use a valid pointer or point `b` to `a`, not `&a`. – this Jan 13 '14 at 08:39
2

just `int *b = a` is enough either use `int *b = &a[0]` – Grijesh Chauhan Jan 13 '14 at 08:40
Remove the cast *and* the ampersand. `int *b = a;` – WhozCraig Jan 13 '14 at 08:41
1

waning message is removed. May be best is using proper pointer. But why cast is wrong here !!!!! – Jeyaram Jan 13 '14 at 08:43
@self. why cast is incorrect? it is not wrong. – Grijesh Chauhan Jan 13 '14 at 08:43
The code will ( probably always )work, but that doesn't mean it is correct. – this Jan 13 '14 at 08:45
@self. it is not the way we do but it is correct and cast is correct. – Grijesh Chauhan Jan 13 '14 at 08:47
ty, but still i cant understand why b and a value's are equal. b is a pointer to a and a is a pointer to a[0] or something else?! – amfad33 Jan 13 '14 at 08:50
The cast is not necessary, you could use this definition of `b`: `int (*b)[2]=&a;`. – Lee Duhem Jan 13 '14 at 08:51
1

@GrijeshChauhan Why use a forced cast to fix an incorrect assign, when you can do things correctly??! – this Jan 13 '14 at 08:52
@self, this casting is common in such cases like handling a large buffer using a pointer. There is no wrong in that. – Jeyaram Jan 13 '14 at 08:53
1

@amfad33 Read answers I linked `a` is of type `int[2]` that decays int o `int*` when assigned to `b`. And `&a` is address of array not first element. to assigned address of array you need to declare b as `int (*b)[2] = &a;` – Grijesh Chauhan Jan 13 '14 at 08:53
1

@self. yes your argument is correct. And it is not correct approach. – Grijesh Chauhan Jan 13 '14 at 08:54
oh! i got that answer ty; :) – amfad33 Jan 13 '14 at 08:54

why in array a[] , &a and a are the same?

1 Answers1