使用cout将数组和char打印到屏幕上时出现意外结果 [英] Unexpected results when printing array and char to screen with cout

问题描述

作为学习C ++的初学者，我试图理解类型为char的数组和类型为int的数组之间的区别.这是我的代码:

As a beginner of learning C++, I am trying to understand the difference between an array of type char and an array of type int. Here is my code:

void IntArray () {
    int array[5] = {5,6,7,8,9};

    cout << "Print int array: " << array << endl;
    cout << "Print int array[0]: " << array[0] << endl;
    cout << "Print int array[0]+1: " << array[0]+1 << endl;
}

void CharArray () {
    char array[5] = {'a', 'b', 'c', 'd', '\0'};

    cout << "Print char array: " << array << endl;
    cout << "Print char array[0]: " << array[0] << endl;
    cout << "Print char array[0]+1: " << array[0]+1 << endl;
}

这是输出:

Print int array: 0xbfd66a88
Print int array[0]: 5
Print int array[0]+1: 6
Print char array: abcd
Print char array[0]: a
Print char array[0]+1: 98

我的问题是:

1. 以下为什么输出字符串'0xbfd66a88'?我期望它返回数组中第一个元素的地址:

1. Why does the following output the string '0xbfd66a88'? I was expecting it to return the address of the first element in the array:

cout << "Print char array: " << array << endl;

为什么以下输出为"98"?我期望它输出字母"b":

Why does the following output '98'? I was expecting it to output the letter 'b':

cout << "Print char array[0]+1: " << array[0]+1 << endl;

推荐答案

1.

因为char数组在流式传输到cout时与其他数组有不同的处理方式，所以<<运算符对于const char*重载.这是为了与C兼容，因此将以空终止的char数组视为字符串.

1.

Because char arrays are treated differently to other arrays when you stream them to cout - the << operator is overloaded for const char*. This is for compatibility with C, so that null-terminated char arrays are treated as strings.

请参见此问题.

这是由于整体促销所致.当您使用char(值为'a')和int(值为1)调用二进制文件+时，编译器会将您的char提升为signed int或unsigned int.哪一个是特定于实现的-取决于char是默认签名还是未签名，并且哪个int可以使用char的整个范围.因此，将使用值'97'和'1'调用+运算符，并返回值'98'.要将其打印为char，您需要先将其转换为

This is due to integral promotion. When you call the binary + with a char (with value 'a') and an int (with value 1), the compiler promotes your char to either a signed int or an unsigned int. Which one is implementation specific - it depends on whether char is signed or unsigned by default, and which int can take the full range of char. So, the + operator is called with the values '97' and '1', and it returns the value '98'. To print that as a char, you need to first cast it:

cout << "Print char array[0]+1: " << static_cast<char>(array[0]+1) << endl;

请参见此问题.

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