Why does printf("%%") print only one percent (%) symbol?

Question

When I run this following code:

#include <stdio.h>
#include <conio.h>

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

I get % as an output?

What might be the reason behind this logic?

Because that's what it does. http://stackoverflow.com/a/1860164/242520 — ta.speot.is, Aug 28 '14 at 06:08
A few quick ways to print both of them: `printf("\%\%");`, `printf("%c%c",'%','%');`, `printf("%s","%%");`. — barak manos, Aug 28 '14 at 06:15
don't vote down for stupidity, the question is clear (very clear). Everyone starts stupid even you. — ctrl-alt-delor, Aug 28 '14 at 06:18
I agree with @richard, don't down vote for stupidity. Down vote for no research effort. — ta.speot.is, Aug 28 '14 at 06:22
@barakmanos the `%` system used by `printf` and family does not interact with the `\ ` system used by the language, so to escape a `%` you have to use a `%` i.e. `%%` and for two `%`s you can do `%%%%`. However it is better to only put formatting information in the formatting string e.g. `printf("%s","%%");` `printf("%s","hello %%"); — ctrl-alt-delor, Aug 28 '14 at 15:50
@richard: Gotcha, thanks for the info. It's actually quite obvious that `'\%'` is **not** a special character, now that I think about it. BTW, you're missing a ` at the end of your comment above. — barak manos, Aug 28 '14 at 15:57

ctrl-alt-delor · Accepted Answer · 2021-05-23T12:09:27.087

That is what printf does: it is print formatted (f for formatted). It uses % as the formatting character. It is the only reserved character and needs to be escaped to represent it self, i.e. %%. See the manual for more information on formatting: printf.

P.S.: Never use a string that is not a part of the program as the first argument. To print a string message that was input by a user, do printf(%s, message);. Otherwise you will have a security hole in your code.

score 2 · Answer 2 · edited May 16 '21 at 20:18

2

% comes into format specifiers.

Example

When we write printf("%d", 20);, it will print 20 rather than %d. because the compiler treats % as a format specifier. In the mind of the compiler, the meaning of % is somewhat special.

So if you want that "%" should be the output, then you must write printf("%%"). Here the first % sign will suppress the meaning of the % format specifier and will print % as an output.

edited May 16 '21 at 20:18

Peter Mortensen

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answered Aug 28 '14 at 06:31

Kailash Karki

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I like your way of explanation, +1 @KailashKarki – Ravi Vaniya Aug 18 '20 at 06:51
Yes, but it is not the compiler that interprets the `%`. It is the `printf` procedure that does it. (```\``` in strings is interpreted by the compiler.) – ctrl-alt-delor May 23 '21 at 12:12

score 1 · Answer 3 · answered Aug 28 '14 at 06:12

From the standard ISO/IEC 9899:1999 (E)

7.19.6.1

Each conversion specification is introduced by the character %.

The conversion specifiers and their meanings are:

% - A % character is written. No argument is converted. The complete conversion specification shall be %%.

score 1 · Answer 4 · edited May 16 '21 at 20:12

1

For C printf, % is a special character which typically indicates a parameter to substitute at that position: printf("Hello, %s\n", "World!"); results in "Hello, World!". There are lots of different things you can put after the % depending on the data you want to output. So that leaves the problem of "What if I want to print a percent symbol"?

The solution: Use %%.

The same is true of the special escape character \. "\n" means to print a new line. If you want to actually print the forward slash, you have to put it twice \\

See Printf format string and MSDN.

edited May 16 '21 at 20:12

Peter Mortensen

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answered Aug 28 '14 at 06:12

Chuck Walbourn

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\ is part of C and C++, % is part of printf, though they work in a similar way. – ctrl-alt-delor Aug 28 '14 at 06:16

Why does printf("%%") print only one percent (%) symbol?

4 Answers4

Example