prints just one backslash?

In C, on mentioning three backslashes, like so:

#include <stdio.h>

int main() {
printf(" ");
}


prints out just one backslash in the output. Why and how does this work?

4 Answers
4

That sequence is:

The C11 draft says (in note 77):

The semantics of these characters were discussed in 5.2.2. If any other
character follows a backslash, the result is not a token and a diagnostic is
required.

On godbolt.org I got:

<source>:8:14: warning: unknown escape sequence ' ' [-Wunknown-escape-sequence]


So you seem to be using a non-conforming compiler, which chooses to implement undefined backslash sequences by just letting the character through.

That is printing:

space
slash
escaped space


The 3rd slash is being interpreted as "slash space"

C11; 6.4.4.4 Character constants:

The double-quote " and question-mark ? are representable either by themselves or by the
escape sequences " and ?, respectively, but the single-quote ' and the backslash
shall be represented, respectively, by the escape sequences ' and .

"

?

"

?

'

'

So, To represent a single backslash, it’s necessary to place double backslashes in the source code. To print two you need four backslash \. In your code extra is a space character, which isn't valid.

\

It is indeed a very simple operation in c. A is just a escape sequences. Hence below statement will print two slash.

c

printf(" \ ");


For example some characters in c are represented with a slash like end of a line character n or end of a string character etc. But if you want to print such a character as it is what will you do? Hence you need to add a escape sequence character in front of it:

c

n

printf("n"); // will print n


But

printf("n"); // will print end of character hence you don't see anything in output


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