C ++解引用数组

本文介绍了C ++解引用数组的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我从来没有读过关于解引用数组的指针一样东西，我认为这不应该工作。但接下来的code不使用Qt创建者和g ++ 4.8的工作：

I never read anything about dereferencing arrays like pointers and I believe it shouldn't work. But the following code does work using QT Creator and g++ 4.8:

int ar[9]{1,2,3,4,5,6,7,8,9}; cout << *ar << endl; //prints the first element of ar

它是正确的行为或只是编译器固定code？

Is it proper behavior or just the compiler fixing the code?

推荐答案

您不能解引用一个数组，只有一个指针。

You cannot dereference an array, only a pointer.

这里发生的事情是，数组类型的前pression，在大多数情况下，隐式转换（衰变）一个指向数组对象的第一个元素。因此， AR 衰变来＆放大器; AR [0] ;提领，让你的AR值[0] ，这是一个 INT 。

What's happening here is that an expression of array type, in most contexts, is implicitly converted to ("decays" to) a pointer to the first element of the array object. So ar "decays" to &ar[0]; dereferencing that gives you the value of ar[0], which is an int.

这最近我的答案讨论这在一些细节的C.用于C ++类似的规则，但C ++有一些更多的情况下，转换不会发生（其中没有发生在你的code）。

This recent answer of mine discusses this in some detail for C. The rules for C++ are similar, but C++ has a few more cases where the conversion does not occur (none of which happen in your code).