Format specifier %02x

Question

I have a simple program :

#include <stdio.h>
int main()
{
        long i = 16843009;
        printf ("%02x \n" ,i);
}

I am using %02x format specifier to get 2 char output, However, the output I am getting is:

1010101 

while I am expecting it to be :01010101 .

Solution 1

%02x means print at least 2 digits, prepend it with 0's if there's less. In your case it's 7 digits, so you get no extra 0 in front.

Also, %x is for int, but you have a long. Try %08lx instead.

Solution 2

%x is a format specifier that format and output the hex value. If you are providing int or long value, it will convert it to hex value.

%02x means if your provided value is less than two digits then 0 will be prepended.

You provided value 16843009 and it has been converted to 1010101 which a hex value.

Solution 3

Your string is wider than your format width of 2. So there's no padding to be done.

Comments

[Rohit Kiran]

I have learnt that 'x' in '%x' refers to hexadecimal format and not int. Is it so?

[aragaer]

x refers to "integer in hexadecimal format" as opposed to d which is "integer in decimal format". Both accept int as a value to be printed. lx and ld both accept long.

[Prashant]

Uh no, printf is not printing the MSB to the right, that would be pretty confusing

[bloodphp]

Besides what eckes said, bit numbering inside of most processors isn't really a thing because you generally deal with a byte at a time. Bit ordering matters only for serialization. Endianness (big or little) is byte ordering, not bit ordering.

[randmin]

For the sake of completenes also read stackoverflow.com/questions/15108932/c-the-x-format-specifie‌​r and cplusplus.com/reference/cstdio/printf

[chux - Reinstate Monica]

"Both accept int as a value to be printed" --> Hmmm, "%x" is for unsigned, not int. Common enough that int "works" though, even if not specified by C.