Printing a % sign in C using 'printf'

Question

I am trying to print a % sign using printf.

I have tried with no luck:

printf("\%");

I'm sure it's really simple, but I've just started C.

It's `%%`. [Read the manual](http://pubs.opengroup.org/onlinepubs/000095399/functions/printf.html). — netcoder, Nov 28 '13 at 20:42
Related: [Why is percentage character not escaped with backslash in C?](http://stackoverflow.com/questions/17811531/why-is-percentage-character-not-escaped-with-backslash-in-c) (though not necessarily duplicate) — Kninnug, Nov 28 '13 at 20:42

score 15 · Accepted Answer · answered Nov 28 '13 at 20:43

15

Use printf("%%"); The backslash is the escape character for C strings; the compiler interprets it. The percent sign is printfs escape character; the printf routine interprets it.

answered Nov 28 '13 at 20:43

M Oehm

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score 5 · Answer 2 · answered Nov 28 '13 at 20:44

5

Here are two ways:

printf("%%\n");
printf("%c\n", '%');

answered Nov 28 '13 at 20:44

Fiddling Bits

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The second isn't really necessary. It's probably slow than first one. – The Mask Nov 28 '13 at 20:44
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@TheMask There's many ways to do the same thing... just trying to demonstrate that to mitchellt – Fiddling Bits Nov 28 '13 at 20:46

score 3 · Answer 3 · answered Nov 28 '13 at 20:43

3

Try by using the following escape sequence:

%%

Check out all escapes usable in *rintf() functions family

answered Nov 28 '13 at 20:43

The Mask

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Printing a % sign in C using 'printf'

3 Answers3

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