How do you escape the % sign when using printf in C?
printf("hello\%"); /* not like this */
views:
745answers:
10
+35
A:
You can escape it by posting a double '%' like this: %%
Using your example:
printf("hello%%");
Escaping '%' sign is only for printf. If you do:
char a[5];
strcpy(a, "%%");
printf("This is a's value: %s\n", a);
It will print: This is a's value: %%
Pablo Santa Cruz
2009-12-07 14:03:32
"printf("hello%%");" is right. But it's not a escape I think. use printf("hello\045");
NjuBee
2009-12-11 03:12:33
@Pablo Santa Cruz: this method to "escape" `%` is specific to `printf`, correct?
Lazer
2010-03-27 12:20:59
@eSKay: correct. Only for printf.
Pablo Santa Cruz
2010-03-27 13:04:52
for formatted functions (afaik they ends all with an `f`) in general, indeed!
ShinTakezou
2010-07-01 08:26:27
This is a special case of the very common rule in escaping systems that to get a literal escape symbol you use <escape symbol><escape symbol>.
dmckee
2010-07-01 14:49:59
+10
A:
If there are no formats in the string, you can use puts (or fputs):
puts("hello%");
if there is a format in the string:
printf("%.2f%%", 53.2);
As noted in the comments, puts appends a \n to the output and fputs does not.
Sinan Ünür
2009-12-07 14:04:59
+1 , so many people forget about (or have never heard of) puts(). I see so many people using printf() for strings that have nothing to expand. This is the best answer.
Tim Post
2009-12-08 05:54:41
Worth mentioning fputs() also, as it directly reciprocates to fprintf().
Tim Post
2009-12-08 05:55:15
puts also appends a newline [even if you already have one]. If you want that, great. Otherwise...
Mikeage
2009-12-08 10:51:15
@Sinan Ünür: thanks for reminding me about `puts`. I never thought of `puts` for printing strings and jumped straight to `printf`. Not anymore.
Lazer
2010-03-24 17:03:12
+13
A:
As others have said, %% will escape the %.
Note, however, that you should never do this:
char c[100];
char *c2;
...
printf(c); /* OR */
printf(c2);
Whenever you have to print a string, always, always, always print it using
printf("%s", c)
to prevent an embedded % from causing problems [memory violations, segfault, etc]
Mikeage
2009-12-07 14:06:41
The warning is generally appropriate however there may be situations in which you want to do "this" - as long as you know that the string you provide will be interpreted as a format string.
PP
2009-12-07 14:17:42
I came up with an alternate solution once - copy the buffer to another buffer and then go through it doubling up the % signs. I eventually came across this idea and replaced a 20-30 line function with one line. Don't worry, I did beat myself severely about the head, as I deserved.
Graeme Perrow
2009-12-07 14:19:53
True, but don't forget that 99% of the time when you get a format string, you get the arguments as a va_list, so you need to use vprintf. Thus, I'm technically correct ;)
Mikeage
2009-12-07 14:20:01
+1
A:
The backslash in C is used to escape characters in strings. Strings would not recognize % as a special character, and therefore no escape would be necessary. Printf is another matter: use %% to print one %.
Ralph Rickenbach
2009-12-07 14:07:08
+1
A:
Like this:
printf("hello%%");
//-----------^^ inside printf, use two percent signs together
Salman A
2009-12-07 14:28:18
+6
A:
Nitpick:
You don't really escape the % in the string that specifies the format for the printf() (and scanf()) family of functions.
The %, in the printf() (and scanf()) family of functions, starts a conversion specification. One of the rules for conversion specification states that a % as a conversion specifier (immediately following the % that started the conversion specification) causes a '%' character to be written with no argument converted.
The string really has 2 '%' characters inside (as opposed to escaping characters: "a\bc" is a string with 3 non null characters; "a%%b" is a string with 4 non null characters).
pmg
2009-12-07 14:28:42
techinically, it is still "escaping"; characters that are special need a way to "escape" their special meaning and be back to their "character nature"
ShinTakezou
2010-07-01 08:29:23
+1
A:
Use this:
#include <stdio.h>
printf("hello%s%s");
A Complete list of format specifiers used with printf can be found here: