# Confusions about (& A + 1)

Source: Internet
Author: User

Take a look at the following programs:

Q:

If int A [5];

What does & A + 1 mean?

A:

& A + 1 does not indicate that the address of a (set to ox0010) is added with 1 and becomes 0x0011. because a is an array of five int types, "+ 1" in "& A + 1" indicates a space equivalent to "1" A size (or an offset ), in this case, & A + 1 indicates a [5].

Q:

If int * PTR = (int *) (& A + 1 );

What does PTR mean?

What does PTR-1 mean?

A:

Since & A + 1 indicates a [5], PTR is a [5].

PTR is int type pointer, so "ptr-1" will subtract "1" int type pointer space, this is a [5-1] = A [4].

// **************** (From the http://sculibin.bokee.com/5628791.html) **************************************

Take a look at the specific procedures below:

Main ()
{
Int A [5] = {1, 2, 3, 4, 5 };
Int * PTR = (int *) (& A + 1 );

Printf ("% d, % d", * (a + 1), * (ptr-1 ));
}
Output: 2, 5

Explanation:
* (A + 1) is a [1], * (ptr-1) is a [4], the execution result is 2, 5
& A + 1 is not the first address + 1. The system will consider that the offset of adding an array a is the offset of an array (in this example, It is 5 Int values)
Int * PTR = (int *) (& A + 1 );
Then PTR is actually & (A [5]), that is, a + 5
The reason is as follows:
& A is an array pointer and its type is int (*) [5];
The pointer plus 1 must add a certain value based on the pointer type,
Different types of pointers + 1 increase with different sizes
A is an int array pointer with a length of 5, so you need to add 5 * sizeof (INT)
Therefore, PTR is actually a [5].
But the PRT and (& A + 1) types are different (this is important)
So the prt-1 will only subtract sizeof (int *)
A, & A has the same address but different meanings. A is the first address of the array, that is, the address of a [0], and a is the first address of the object (array, A + 1 is the address of the next element of the array, that is, a [1]. & A + 1 is the address of the next object, that is, a [5].

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