C Programming Error: Understanding '%%d' vs. '%d' in printf()

There is an error in this code. It should be:

main()
{
   int a=2,c=5;
   printf('a=%d,c=%d\n',a,c);
}

The format specifier %%d should be %d to print the values of the variables a and c.

The error lies in the use of %%d instead of %d within the printf() function. In C, %d is the format specifier used to print integer values. Using %%d would actually result in the literal string %d being printed, not the value of the variable.

Explanation:

• printf() is a standard library function in C used to display output on the console.
• Format Specifiers: %d, %f, %s, etc., are used to define the type of data to be printed.
• %d: Specifies that an integer value should be printed.
• %%d: The double percent sign %% is used to escape the percent sign character itself, resulting in the literal %d being printed, not the integer value.

Corrected Code:

main()
{
   int a=2,c=5;
   printf('a=%d,c=%d\n',a,c);
}

This code will correctly print the following output to the console:

a=2,c=5