C ++ address addition constant

Source: Internet
Author: User

Main (){

Int a [5] = {1, 2, 3, 4, 5 };

Int * ptr = (int *) (& a + 1 );

Printf ("% d, % d", * (a + 1), * (ptr-1 ));


Output: 2, 5

* (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 an array a is the offset of an array (in this example, it is five int values) int * ptr = (int *) (& a + 1 ); the ptr is actually & (a [5]), that is, the reason for a + 5 is as follows: & a is an array pointer and its type is int (*) [5]; the pointer plus 1 should add a certain value according to the pointer type, and the size of the pointer plus 1 of different types will be different. A is an int array pointer with a length of 5, so you must add
5 * sizeof (int) So 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 address is the same, but the meaning is not the same, a is the first address of the array, that is, a [0] address, & a is the first address of the object (array), and 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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