#include <stdio.h>void Main (){int a[5]={1,2,3,4,5};int *ptr= (int *) (&a+1);printf ("%d,%d", * (a+1), * (ptr-1));return;}output is: 2,5If int*ptr = (int*) (&a+1) is converted to int* ptr= (int *) (a+1) then the output is 2,1 Explanation:* (a+1) is actually very simple refers to a[1], the output is 2.The problem is the second point, what is the output of the * (PTR-1)? as explained below, &a+1 is not the first address +1, the system will consider adding an entire array of a, which offsets the size of the entire array a (that is, the size of the 5 int). so int *ptr= (int *) (&a+1), actually PTR is actually & (A[5]), which is a+5.What is the reason? &a represents a pointer to an entire array, its value is the same as the address value of the first element of a group, but the memory size that the pointer points to is the total size of the entire array. So &a+1 means the next memory cell of the last element of the entire array a starts the array pointer and then convert the pointer to a pointer to int, then Ptrl points to the 4 bytes (int*) memory that starts at the next memory cell after the last element of array A. &a is an array pointer with the type int (*) [5], and the pointer plus 1 to add a certain value according to the pointer type, the different type of pointer +1 after the increase in size, a is a length of 5 int array pointer, so to add 5*sizeof (int), so ptr is actually a [5], but the PTR and (&a+1) type is not the same, this is very important, so ptr-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, the address of a[0],& A is the first address of an 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, or A[5]. In addition, there is a similar written question:about int a[10]; Ask the following address which cannot represent a[1]?
A. a+sizeof (int)
B. &a[0]+1
C. (int*) &a+1
D. (int*) ((char*) &a+sizeof (int.))answer: A. a+sizeof (int) parsing:
A. a+sizeof (int)//incorrect, equivalent to pointer operation A + 4 on a 32-bit machine
B. &a[0]+1//correct, array first element address plus 1, according to pointer operation is a[1] address
C. (int*) &a+1//correct, the array address is coerced to int*, then add 1, so and b means a meaning
D. (int*) ((char*) &a+sizeof (int))//correct, the data address is first converted to char*, then add 4, according to the pointer operation formula, move forward 4 * sizeof (char), then be converted to int*, obviously a[1] The AddressThe ps:c Answer (int*) &a+1 is not the same as the (int *) (&a+1) in the above topic, (int*) &a+1 takes the address of the array and then forces the address of the group's first element to be added 1, so (int*) &a+1 can represent a[1] addresses, while (int *) (&a+1) refers to the address of the next array based on the above analysis.
The difference between an array pointer and a pointer to the first element of a group
