c++ - 为什么语句 “cout << '\\\\';” 不会失败？

Question

源代码如下。

cout << '\\' << endl;  //OK, output is \  
cout << '\\\\' << endl;  //OK, output is an integer 23644, but why? 

声明cout << '\\\\' << endl;调用类 ostream 的以下函数.

_Myt& __CLR_OR_THIS_CALL operator<<(int _Val)

我知道写'\\\\'这个表达式很奇怪, 但我不明白为什么它不会失败。结果如何解释？

Answer 1

这是一个多字 rune 字，类型为 int。

[lex.ccon]/2 :

An ordinary character literal that contains more than one c-char is a multicharacter literal. A multicharacter literal, or an ordinary character literal containing a single c-char not representable in the execution character set, is conditionally-supported, has type int, and has an implementation-defined value.

你应该使用"\\\\"，也就是char const[3]:两个\和最后一个NUL字节。