ostream cout和char * [英] ostream cout and char *

问题描述

我有一个这样的字符数组：

I have an array of chars like this one:

char arr[3]="hi";
cout << arr;// this will print out hi

so是运算符<有一个重载版本，它接受一个ostream对象和char *。因此 cout<< arr; 第一个arr将衰减到聊天*。然后操作符<<（）将打印出char指针指向的东西，直到它找到空字符为止。

So is the operator<< has an overloaded version that takes an ostream object and char *. so cout<<arr; first arr will decays to a chat * . and then operator<<() will print out what the char pointer is pointing to until it find a null-character ?

推荐答案

您的 ostream 和 istream > operator 和 operator>> 重载以获取 char * 并且阵列衰减到指向第一元素的指针。所以，是的，它做你所说的。

Your ostream and istream do have operator<< and operator>> overloaded to take a char*, and arrays decay into pointers to the first element. So, yes it does what you say it does.

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