为什么声明“cout <<'\\\\\\\\';”没有失败？

Question

The source code is as the following.

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

The statement cout << '\\\\\\\\' << endl; invokes the following function of class ostream .

_Myt& __CLR_OR_THIS_CALL operator<<(int _Val)

I know it is strange to write the expression '\\\\\\\\' , But I don't understand why it doesn't fail. How to explain the result?

Answer 1

This is a multicharacter literal and has type 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.

You should use "\\\\\\\\" , which is char const[3] : two \\ and a NUL byte at the end.