什么是之间和QUOT的差异;为int * A [5] QUOT;和INT（*一）[5]＆QUOT ;? [英] What's the difference between "int *a[5]" and int(*a)[5]"?

问题描述

他们会在不同的C和C ++的工作？

P.S。我的第一个问题，我是一个编程小白所以请尽量保持asnwer基本和简单的：）

感谢你在前进！

解决方案

1. 为int * A [5] -
   这意味着，a是一个指针数组，即阵列中的每个部件a是一个指针搜索
   整数类型;数组的每个成员都可以容纳一个整数的地址。




2. INT（*一）[5] -
   这里a是一个指针5的整数的数组，换言之
   一指向包含5整数数组。

例如：

 ＃包括LT＆;＆stdio.h中GT;
   诠释的main（）
   {
           INT B = 3;
           INT C = 4;
           为int * a [2] = {和B，和C}; //是一样的---为int * a [] = {和B，和C}
           的printf（价值指向一个[0] =％d个，由[1] =％d个\\ N，* A [0]，*一[1]）;
           返回0;
   }
 

Would they work differently on C and C++?

P.s. My first question and I am a programming noob so please try to keep the asnwer basic and simple :)

Thank you in advance !

解决方案

1. int *a[5] - It means that "a" is an array of pointers i.e. each member in the array "a" is a pointer
   of type integer; Each member of the array can hold the address of an integer.

2. int (*a)[5] - Here "a" is a pointer to the array of 5 integers, in other words "a" points to an array that holds 5 integers.

Example :

#include<stdio.h>
   int main()
   {
           int b = 3;
           int c = 4;
           int *a[2] = {&b, &c}; // is same as ---int *a[] = {&b, &c} 
           printf("value pointed to by a[0] = %d, by a[1] = %d\n", *a[0], *a[1]);
           return 0;
   }

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