问题指针数组的声明 [英] Problem with declaration of an array of pointers

问题描述

当我执行此code

#include<stdio.h>

int main() {
 int (*x)[5];
printf("\nx = %u\nx+1 = %u\n&x = %u\n&x + 1 = %u",x,x+1,&x,&x+1);
}

这是用C或C ++的输出：

This is the output in C or C++:

x = 134513520
x+1 = 134513540
&x = 3221191940
&x + 1 = 3221191944

请解释一下。此外之间有什么区别：

Please explain. Also what is the difference between:

INT X [5] 和 INT（* X）[5] ？

推荐答案

• INT X [5] 5整数数组


• INT（* X）[5] 是指针的5个整数数组

• int x[5] is an array of 5 integers
• int (*x)[5] is a pointer to an array of 5 integers

当你增加一个指针，您可以通过所指向的类型的大小递增。 X + 1 因此， 5 *的sizeof（INT）字节不仅仅是放大X - 给 8048370 和 8048384 的十六进制值与0x14的，或20的差异

When you increment a pointer, you increment by the size of the pointed to type. x+1 is therefore 5*sizeof(int) bytes larger than just x - giving the 8048370 and 8048384 hex values with a difference of 0x14, or 20.

＆放大器; X 是一个指针的指针 - 所以，当你增加它添加的sizeof（指针）字节 - 这给 bf9b08b4 和 bf9b08b8 的十六进制值，为4的区别

&x is a pointer to a pointer - so when you increment it you add sizeof(a pointer) bytes - this gives the bf9b08b4 and bf9b08b8 hex values, with a difference of 4.

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