Format Specifier %02x - Stack Overflow

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Learn more about Labs Format specifier %02x Ask Question Asked 11 years, 3 months ago Modified 5 years, 5 months ago Viewed 192k times 52

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 .

• c
• printf
• format-specifiers

Share Improve this question Follow edited May 17, 2017 at 8:55 Mini Bhati 34355 silver badges2020 bronze badges asked Aug 26, 2013 at 7:35 user2717225user2717225 53111 gold badge44 silver badges55 bronze badges 0 Add a comment |

4 Answers 4

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 66

%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.

Share Improve this answer Follow edited Aug 26, 2013 at 9:22 Chris Young 15.8k77 gold badges3838 silver badges4242 bronze badges answered Aug 26, 2013 at 7:38 aragaeraragaer 17.8k66 gold badges5050 silver badges5252 bronze badges 4

• 4 I have learnt that 'x' in '%x' refers to hexadecimal format and not int. Is it so? – Rohit Kiran Commented Jan 21, 2015 at 13:41
• 18 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. – aragaer Commented Jan 22, 2015 at 0:28
• 1 For the sake of completenes also read stackoverflow.com/questions/15108932/c-the-x-format-specifier and cplusplus.com/reference/cstdio/printf – randmin Commented Mar 22, 2018 at 23:12
• "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. – chux Commented Jun 23, 2021 at 18:17

Add a comment | 18

%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.

Share Improve this answer Follow edited Jun 21, 2019 at 8:48 leiyc 9431111 silver badges2323 bronze badges answered Nov 3, 2017 at 6:13 PulkitRajputPulkitRajput 66977 silver badges88 bronze badges Add a comment | 2

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

Share Improve this answer Follow answered Aug 26, 2013 at 7:37 PP.PP. 10.9k77 gold badges4747 silver badges5959 bronze badges Add a comment | -3

You are actually getting the correct value out.

The way your x86 (compatible) processor stores data like this, is in Little Endian order, meaning that, the MSB is last in your output.

So, given your output:

10101010

the last two hex values 10 are the Most Significant Byte (2 hex digits = 1 byte = 8 bits (for (possibly unnecessary) clarification).

So, by reversing the memory storage order of the bytes, your value is actually: 01010101.

Hope that clears it up!

Share Improve this answer Follow edited Feb 19, 2016 at 19:54 Ahmed Akhtar 1,46311 gold badge1717 silver badges2929 bronze badges answered Feb 19, 2016 at 18:24 RJMRJM 7144 bronze badges 2

• Uh no, printf is not printing the MSB to the right, that would be pretty confusing – eckes Commented May 17, 2017 at 8:59
• 3 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. – Daniel Papasian Commented Aug 6, 2017 at 16:09

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