How to print % using printf()?
Asked by Tanuj
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.
The flag character
The field width
The precision
The length modifier
The conversion specifier:
See http://swoolley.org/man.cgi/3/printf 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 “%%”
/* Program to print %*/#include<stdio.h>/* Program to print %*/int main(){ printf("%%"); getchar(); return 0;} |
We can also print “%” using below.
printf("%c", '%');printf("%s", "%"); |
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