How to escape the % (percent) sign in C's printf?

Question

How do you escape the % sign when using printf in C?

printf(\"hello\\%\"); /* not like this */

Answer 1

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]

Answer 2

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 %.

Answer 3

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*%"

Answer 4

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: %%

Answer 5

You can simply use % twice, that is "%%"

Example:

printf("You gave me 12.3 %% of profit");

Answer 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).