Why does the statement “cout << '\\\\';” not fail?

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?

What do you mean by "fail"? Do you think it should fail to compile? If so, with what error? It's not clear why you find this behavior unexpected. — David Schwartz, Apr 08 '16 at 09:24
See http://stackoverflow.com/q/3960954/2491746 for more discussion. — Simple, Apr 08 '16 at 09:42

score 15 · Answer 1 · edited Apr 08 '16 at 09:30

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.

Why does the statement “cout << '\\\\';” not fail?

1 Answers1