指针数组C ++ [英] pointer to array c++

问题描述

什么是以下code干什么？

  INT政[] = {9,8};
INT（* j）条=克;
 

从我的理解它创建一个指向2整型数组。
但后来为什么这项工作：

  INT X =Ĵ[0];
 

这不工作：

  INT X =（* j）条[0];
 

解决方案

括号是在你的例子是多余的。指针不关心是否有涉及到一个数组 - 只知道它指向一个 INT

  INT政[] = {9,8};
  INT（* j）条=克;
 

也可以被改写为

  INT政[] = {9,8};
  为int * J =克;
 

这也可被改写为

  INT政[] = {9,8};
  为int * J =＆放克[0];
 

一个指针到一个数组看起来像

  INT政[] = {9,8};
  INT（* j）条[2] =＆放大器;克;  //提领'J'和访问数组元素为零
  INT N =（* j）条[0];
 

有一个关于指针声明（以及如何神交他们）在这里这个链接一个良好的阅读：<一href=\"http://www.$c$cproject.com/Articles/7042/How-to-inter$p$pt-complex-C-C-declarations\">http://www.$c$cproject.com/Articles/7042/How-to-inter$p$pt-complex-C-C-declarations

What is the following code doing?

int g[] = {9,8};
int (*j) = g;

From my understanding its creating a pointer to an array of 2 ints. But then why does this work:

int x = j[0];

and this not work:

int x = (*j)[0];

解决方案

The parenthesis are superfluous in your example. The pointer doesn't care whether there's an array involved - it only knows that its pointing to an int

  int g[] = {9,8};
  int (*j) = g;

could also be rewritten as

  int g[] = {9,8};
  int *j = g;

which could also be rewritten as

  int g[] = {9,8};
  int *j = &g[0];

a pointer-to-an-array would look like

  int g[] = {9,8};
  int (*j)[2] = &g;

  //Dereference 'j' and access array element zero
  int n = (*j)[0];

There's a good read on pointer declarations (and how to grok them) at this link here: http://www.codeproject.com/Articles/7042/How-to-interpret-complex-C-C-declarations

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