cpp cout中的单引号

Question

摘自《编程：使用 C++ 的原则与实践》第 72–73 页：

We saw that we couldn’t directly add chars or compare a double to an int. However, C++ provides an indirect way to do both. When needed, a char is converted to an int and an int is converted to a double. For example:

char c = 'x'; 
int i1 = c; 
int i2 = 'x';

Here both i1 and i2 get the value 120, which is the integer value of the character 'x' in the most popular 8-bit character set, ASCII. This is a simple and safe way of getting the numeric representation of a character. We call this char-to-int conversion safe because no information is lost; that is, we can copy the resulting int back into a char and get the original value:

char c2 = i1; 
cout << c << ' << i1 << ' << c2 << '\n';

This will print x 120 x

我不明白这里使用的单引号。当我尝试时，它打印 x540818464x.

Answer 1

您应该试试这段代码（如下）。你或那本书打错字了。

#include <iostream>
using namespace std;

int main() {
    char c = 'x';
    int i1 = c;
    int i2 = 'x';
    char c2 = i1;
    std::cout << c << ' ' << i1 << ' ' << c2 << '\n';
    return 0;
}

结果：

Answer 2

对该代码的唯一解释是这本书印刷错误，或者这是一个明显的拼写错误。正确的代码应该是

cout << c << ' ' << i1 << ' ' << c2 << '\n';

Answer 3

' << i1 << ' 是一个多字符文字，具有类型 int 和实现定义的值。

你可能想要：cout << c << ' ' << i1 << ' ' << c2 << '\n';

使用正则字符 space.

Answer 4

这看起来确实像一个错字，我们可以安全地推测作者的意思：

cout << c << ' ' << i1 << ' ' << c2 << '\n';

不正确的代码虽然可以编译，但是没有意义：

cout << c << ' << i1 << ' << c2 << '\n';

在这里，' << i1 << '（注意单引号）被称为多字符文字，它具有类型int和实现定义的值。

From [lex.ccon]/2:

An ordinary character literal that contains more than one c-char is a multicharacter literal. A multicharacter literal [...] is conditionally-supported, has type int, and has an implementation-defined value.

它的用法相对较少，我个人认为它是一种定义任意常量的方法，例如

enum state { wait = 'wait', start = 'start', /*...*/ };

Answer 5

通过使用单引号，我们可以打印ascii值。所以打印了整个表达式的 ascii 值。

谢谢我希望这会有所帮助

cpp cout中的单引号

Single Quote in cpp cout

c++

cout