The instruction
printf("%4.2f", num);displays four digits including two decimal places?
No, that is not how it works.
A format string is specified in the following format:
%[flags][width][.precision][size]type
Specifiers in [] are optional.
The width specifier indicates the minimum number of characters to output total (it may be more!). If not specified, the width defaults to 1.
The precision specifier indicates how many digits the f type prints after a decimal point (the precision has other meanings for different types). If not specified, the precision defaults to 6 for the f type.
In your example format string: %4.2f, 4 is the width, 2 is the precision, and f is the type.
So, for example, given double num = 123.0;:
printf("%f", num);
Has a width of 1 and a precision of 6, so it prints "123.000000" – at least 1 character, including the decimal and 6 digits following the decimal.
printf("%4.2f", num);
Has a width of 4 and a precision of 2, so it prints "123.00" – at least 4 characters, including the decimal and 2 digits following the decimal.
printf("%10.2f", num);
Has a width of 10 and a precision of 2, so it prints " 123.00" – at least 10 characters, including 4 leading spaces added as padding, the decimal, and 2 digits following the decimal.
Live Demo
Refer to this printf reference for more details about the format string and its inputs and outputs.