I am trying to print hexadecimal values in C.
A simple known program to print hexadecimal value...( Using %x or %X format specifier)
#include<stdio.h>
int main()
{
unsigned int num = 10;
printf("hexadecimal value of %u is %x\n", num, num);
}
Output: hexadecimal value of 10 is a
Instead of getting output as a, I need to get output as 0xa in hexadecimal format (making if more explicit for the user that the value in expresses in hexadecimal format).
There are 2 ways to achieve this:
using 0x%x, which ensures even 0 is printed as 0x0. 0x%X is the only solution to print the x in lowercase and the number with uppercase hexadecimal digits. Note however that you cannot use the width prefix as the padding spaces will appear between the 0x and the digits: printf(">0x%4x<", 100) will output >0x 64<
using %#x, which adds the 0x prefix if the number is non zero and is compatible with the width prefix: printf(">%#6x<", 100) will output > 0x64<, but both printf(">%#06x<", 100) and printf(">0x%04x<", 100) will output >0x0064<.
The other answers have explained well how the "0x" prefix can be obtained.
I would like to add that hexadecimal numbers are often used with a fixed width (e.g. 2 or 4 digits). This can be obtained as follows:
printf("0x%04x", num);
which will print e.g.
0x0054
%#xalso works: see en.cppreference.com/w/c/io/fprintfo, u, x, Xare documented. It says The unsigned int argument is converted to unsigned octal (o), unsigned decimal (u), or unsigned hexadecimal (x and X) notation. The#modifier is documented above that, under "flag characters".%dexpects a signedint. You could use%uinstead of%d, which prints an unsigned decimal value. So you could dounsigned int num = 10;thenprintf("%u %#x\n", num, num);That would be strictly correct, since both format specifiers expect anunsigned int.