printing % in the output

Question

why do I need to write two % in this statement to have one in the output?
printf("Top 10%%.\n");

and if I write one % I don't have it in the output!

Answer 1

Preamble:

In C, % character has a special meaning when being used in I/O functions (e.g. printf()) as part of a string literal. Being a prefix indicating the beginning of a conversion specifier, after that, a sequence of other characters* is followed which is needed to set the conversion behavior to what is necessary. The fields of a conversion specifier shown below:

"% [flag] [length modifier] [format specifier]"

► flag: https://port70.net/~nsz/c/c11/n1570.html#7.21.6.1p6

► length modifier: https://port70.net/~nsz/c/c11/n1570.html#7.21.6.1p7

► format specifier: https://port70.net/~nsz/c/c11/n1570.html#7.21.6.1p8

Note that, in the case of an input function such as `scanf()` there's no flag field.
~~~~~~~~~~~~~~
Answer to the question:

Given how the language has defined this feature, it also provided the way by which the actual % character can be interpreted as to its literal form. Pretty much the same rationale when dealing with \ (backslash) character.

"\" Error: \" escapes " (double quote) character
"\"" prints "
"\\" prints \

"%" Warning: spurious trailing ‘%’ in format
"%%" prints %

________
* 1) flag 2) length modifier 3) format specifier — format specifier is mandatory among the three.