How to display hexadecimal numbers in C

Question

I have a list of numbers as below      0  16  32  48       I need to output those numbers in hexadecimal as      0000 0010 0020 0030 0040       I have tried solution such as   printf    4x  a      where a is an integer   but the result that I got is      0000  0001  0002  0003  0004       I think I m close there  Can anybody help as I m not so good at printf in C   Thanks

User · Accepted Answer

Try   printf   04x  a      0 - Left-pads the number with zeroes  0  instead of spaces  where padding is specified  4  width  - Minimum number of characters to be printed  If the value to be printed is shorter than this number  the result is right justified  within this width by padding on the left  with the pad character   By default this is  a blank space  but the leading zero we used  specifies a zero as the pad char  The value is not truncated even if the result is larger  x - Specifier for hexadecimal integer    More here

User · Answer

i use it like this   printf  my number is 0x 02X n  number      output  my number is 0x4A   Just change number  2  to any number of chars You want to print

User · Answer

You can use the following snippet code    include lt stdio h gt  int main int argc  char  argv         unsigned int i      printf  decimal  hexadecimal n        for  i   0  i  lt   256  i  16          printf   04d     0x 04X n   i  i       return 0      It prints both decimal and hexadecimal numbers in 4 places with zero padding

User · Answer

Your code has no problem  It does print the way you want   Alternatively  you can do this   printf   04x  a

[c] How to display hexadecimal numbers in C?

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