How to print % using printf()?
Here is the standard prototype of printf function in C:
int printf(const char *format, ...);
The format string is composed of zero or more directives: ordinary characters (not %), which are copied unchanged to the output stream; and conversion specifications, each of argument (and it is an error if insufficiently many arguments are given).
The character % is followed by one of the following characters.
- Flag character
- Field width
- Length modifier
- Conversion specifier
Check this for details of all the above characters. The main thing to note in the standard is the below line about conversion specifier.
A '%' is written. No argument is converted. The complete conversion specification is'%%'.
So we can print “%” using “%%”
We can also print “%” using below.
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