When I going to print a hexadecimal value using printf it takes the input as decimal value why?

Question

For example

printf("%x",16);

This printf prints 10 instead of 16 why it is not take the value as hexadecimal pls someone explain this

Answer 1

16 is a decimal integer constant. Thus the hexadecimal representation of the constant is 10. If you want to specify a hexadecimal integer constant then write 0x16u. Pay attention to that the conversion specifier x expects an argument of the type unsigned int. So the suffix u is used in the hexadecimal integer constant.

So the call of printf can look like

printf( "%x\n", 0x16u );

and the output will be 16.

Or

printf( "%#x\n", 0x16u );

and the output will be 0x16.

Answer 2

When you pass a number to a function, you are not passing decimal or hexadecimal. Conceptually, you are passing a value, an abstract mathematical number.

That number is represented in some type, and C uses binary to represent values in integer types (along with some supplement of binary for signed numbers, usually two’s complement).

Whether you write printf("%x\n", 16u) or printf("%x\n", 0x10u);, the compiler converts that numeral in source code, 16u or 0x10u, to an unsigned int value, represented in binary. It is that resulting value that is passed to printf, not a decimal “16” or a hexadecimal “10”.

(I use 16u and 0x10u rather than 16 or 0x10 because printf expects an unsigned int, not an int.)

The %x directive tells printf to expect an unsigned int value and to convert it to a hexadecimal numeral. So the original input form is irrelevant; %x means to produce hexadecimal regardless of the original form.