How do I print the percent sign(%) in C?

Question

Why doesn't this program print the % sign?

#include <stdio.h>

main()
{
     printf("%");
     getch();
}

So did you read the documentation? What did it say about the percent sign? — , Jul 21 '13 at 17:20
This question can be easily answered through Google. The OP could have used his logic, ie (`'//`) is the character constant for `/` and applied this. Isn't there also a man page on `printf`? — TheBlueCat, Jul 21 '13 at 20:38
Perhaps you meant to say that `'\\'` is the character constant for `\ `? The character constant for `/` is `'/'`. — This isn't my real name, Jul 22 '13 at 18:57

score 94 · Accepted Answer · edited Jul 21 '13 at 18:08

94

Your problem is that you have to change:

printf("%");

to

printf("%%");

Or you could use ASCII code and write:

printf("%c", 37);

:)

edited Jul 21 '13 at 18:08

Grijesh Chauhan

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answered Jul 21 '13 at 17:14

C_Intermediate_Learner

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C doesn't require ASCII. `printf("%c", 37);` is not portable. – Jul 21 '13 at 18:05
2

@H2CO3 yes but I like the answer because he given an extra trick! – Grijesh Chauhan Jul 21 '13 at 18:09
@GrijeshChauhan I didn't say I didn't like the answer, it's a good answer. I only pointed out that the aforementioned snippet is not portable. – Jul 21 '13 at 18:17
@H2CO3 yup i know...I just like to say you that I might simply post `printf("%%");` answer but not using ascii. You commented here when I was editing this answer. :) – Grijesh Chauhan Jul 21 '13 at 18:22
1

@GrijeshChauhan Of course. No offense :) – Jul 21 '13 at 18:27
5

Why not use `'%'` instead of `37`? – Diti Mar 22 '14 at 16:52

score 27 · Answer 2 · edited May 16 '21 at 19:47

There's no explanation in this topic why to print a percentage sign. One must type %% and not for example an escape character with percentage - \%.

From comp.lang.c FAQ list · Question 12.6:

The reason it's tricky to print % signs with printf is that % is essentially printf's escape character. Whenever printf sees a %, it expects it to be followed by a character telling it what to do next. The two-character sequence %% is defined to print a single %.

To understand why % can't work, remember that the backslash \ is the compiler's escape character, and controls how the compiler interprets source code characters at compile time. In this case, however, we want to control how printf interprets its format string at run-time. As far as the compiler is concerned, the escape sequence % is undefined, and probably results in a single % character. It would be unlikely for both the \ and the % to make it through to printf, even if printf were prepared to treat the \ specially.

So the reason why one must type printf("%%"); to print a single % is that's what is defined in the printf function. % is an escape character of printf's, and \ of the compiler.

Thank you for offering the reason. – Dave Voyles Mar 13 '16 at 16:02 — Dave Voyles, Mar 13 '16 at 16:02

score 8 · Answer 3 · answered Jul 21 '13 at 17:13

8

Use "%%". The man page describes this requirement:

% A '%' is written. No argument is converted. The complete conversion specification is '%%'.

answered Jul 21 '13 at 17:13

Carl Norum

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score 3 · Answer 4 · answered Jul 21 '13 at 17:52

3

Try printing out this way

printf("%%");

answered Jul 21 '13 at 17:52

Santhosh Pai

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This answer could be more useful than that. It doesn't give any reference enabling the readers to learn about their mistakes. – Diti Mar 22 '14 at 16:54

How do I print the percent sign(%) in C?

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