This question already has answers here:
How to escape the % (percent) sign in C's printf
(13 answers)
Closed 7 years ago.
Why doesn't this program print the % sign?
#include <stdio.h>
main()
{
printf("%");
getch();
}
Your problem is that you have to change:
printf("%");
to
printf("%%");
Or you could use ASCII code and write:
printf("%c", 37);
:)
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.
Use "%%". The man page describes this requirement:
% A '%' is written. No argument is converted. The complete conversion specification is '%%'.
Try printing out this way
printf("%%");
Related
This question already has answers here:
How to escape the % (percent) sign in C's printf
(13 answers)
Closed 8 years ago.
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?
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.
% 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.
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 %%.
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.
This question already has answers here:
Why is percentage character not escaped with backslash in C?
(4 answers)
How to escape the % (percent) sign in C's printf
(13 answers)
Closed 9 years ago.
After reading over some K&R C I saw that printf can "recognize %% for itself" I tested this and it printed out "%", I then tried "\%" which also printed "%".
So, is there any difference?
Edit for code request:
#include <stdio.h>
int main()
{
printf("%%\n");
printf("\%\n");
return 0;
}
Output:
%
%
Compiled with GCC using -o
GCC version: gcc (SUSE Linux) 4.8.1 20130909 [gcc-4_8-branch revision 202388]
%% is not a C escape sequence, but a printf formatter acting like an escape for its own special character.
\% is illegal because it has the syntax of a C escape sequence, but no defined meaning. Escape sequences besides the few listed as standard are compiler-specific. In all likelihood the compiler ignored the backslash, and printf did not see any backslash at runtime. If it had, it would have printed the backslash in the output, because backslash is not special to printf.
Both are not the same. The second one will print %, but in case of the first one, you will get compiler warning:
[Warning] unknown escape sequence: '%' [enabled by default]
The warning is self explanatory that there is no escape sequence like \% in C.
6.4.4.4 Character constants;
says
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 \\.
It is clear that % can't be represented as \%. There isn't any \% in C.
When "%%" is passed to printf it will print % to standard output, but "\%" in not an valid escape sequence in C. Hence the program will compile, but it will not print anything and will generate a warning:
warning: spurious trailing ‘%’ in format [-Wformat=] printf("%");
The list of escape sequences in C can be found in Escape sequences in C.
This won't print % for the second printf.
int main()
{
printf("%%\n");
printf("\%");
printf("\n");
return 0;
}
Output:
%
This question already has answers here:
How to escape the % (percent) sign in C's printf
(13 answers)
Closed 9 years ago.
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.
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.
Here are two ways:
printf("%%\n");
printf("%c\n", '%');
Try by using the following escape sequence:
%%
Check out all escapes usable in *rintf() functions family
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
How to escape the % (percent) sign in C's printf
How can I print '%' in C?
E.g.:
printf("A: %.2f%", pc);
...fails, the compiler complains there is an invalid conversion. Of course an easy way is;
printf("A: %.2f%c", pc, '%');
But it's rather inelegant...
I looked on the web, but I didn't find any escape sequence for %. I thought % would work, but it doesn't.
printf("A: %.2f%%", pc);
Just double the '%' in the format string and it will print '%'.
For future printf reference, type:
man 3 printf
on any Linux command prompt. It can do a lot of crazy stuff that most people just aren't aware of.
%% will output % using printf.
Check the example at the end of page here
How do you escape the % sign when using printf in C?
printf("hello\%"); /* not like this */
You can escape it by posting a double '%' like this: %%
Using your example:
printf("hello%%");
Escaping the '%' 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: %%
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, segmentation faults, etc.).
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.
With itself...
printf("hello%%"); /* like this */
Use a double %%:
printf("hello%%");
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).
Like this:
printf("hello%%");
//-----------^^ inside printf, use two percent signs together
You can use %%:
printf("100%%");
The result is:
100%
You are using the incorrect format specifier. You should use %% for printing %. Your code should be:
printf("hello%%");
Read more all format specifiers used in C.
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 %.
You can simply use % twice, that is "%%"
Example:
printf("You gave me 12.3 %% of profit");
Yup, use printf("hello%%"); and it's done.
The double '%' works also in ".Format(…).
Example (with iDrawApertureMask == 87, fCornerRadMask == 0.05):
csCurrentLine.Format("\%ADD%2d%C,%6.4f*\%",iDrawApertureMask,fCornerRadMask) ;
gives the desired and expected value of (string contents in) csCurrentLine;
"%ADD87C, 0.0500*%"