问题
How do you escape the % sign when using printf in C?
printf(\"hello\\%\"); /* not like this */
回答1:
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: %%
回答2:
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]
回答3:
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.
回答4:
With itself...
printf("hello%%"); /* like this */
回答5:
use a double %%
回答6:
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).
回答7:
Like this:
printf("hello%%");
//-----------^^ inside printf, use two percent signs together
回答8:
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 %.
回答9:
Yup, use printf("hello%%"); and it's done.
回答10:
You can use %%:
printf("100%%");
The result is:
100%
回答11:
you are using incorrect format specifier you should use %% for printing %. Your code should be:
printf("hello%%");
Read more all format specifiers used in C.
回答12:
You can simply use % twice, that is "%%"
Example:
printf("You gave me 12.3 %% of profit");
回答13:
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*%"
来源：https://stackoverflow.com/questions/1860159/how-to-escape-the-percent-sign-in-cs-printf