Why putchar('\\\'); won't work

Question

I want to get the output \\ ,so I type this to my code putchar('\\\'); but it doesn't work out ,and I get the following error

D:\c++ work\change\change.cpp(17) : error C2001: newline in constant
D:\c++ work\change\change.cpp(17) : fatal error C1057: unexpected end of file in macro expansion

why?

Are you asking about 2 escape-sequences and an unterminated multi-character-constant, or a single escape-sequence in a normal character-constant? — Deduplicator, Nov 12 '14 at 13:31

Borgleader · Answer 1 · 2014-11-12T13:38:44.863

5

\\ is an escaped \, then you have \' which is an escaped '. I'm guessing you meant to write: putchar('\\'); or putchar('\\\\'); (the latter being incorrect, you should do the former twice or use another function)

As you have it now (\\\) you're escaping the last ' which is causing your compiler to "misinterpret" your code. (I put misinterpret in quotes because it's actually interpreting it correctly, it's just not immediately obvious).

tl;dr

putchar('\\');
putchar('\\');

or

puts("\\\\");

will get you the correct output. (See live example)

edited Nov 12 '14 at 13:38

answered Nov 12 '14 at 13:31

Borgleader

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@Deduplicator yes i should clear that up – Borgleader Nov 12 '14 at 13:33
@Deduplicator `'\\\\'` is actually [implementation-defined](http://stackoverflow.com/a/19394188/1009479). – Yu Hao Nov 12 '14 at 13:42
@YuHao: You mean the value of `\\\\`. Yes. But it won't be the same as for a character constant... (In theory it could, yes again) – Deduplicator Nov 12 '14 at 13:44
@Deduplicator My point is, it's not undefined behavior, just implementation-defined. – Yu Hao Nov 12 '14 at 13:45

Why putchar('\\\'); won't work

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