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 */

Solution 1

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: %%

Solution 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, segmentation faults, etc.).

Solution 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.

Solution 4

With itself...

printf("hello%%"); /* like this */

Solution 5

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).

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