How do you format an unsigned long long int using printf

Question

include  lt stdio h gt  int main         unsigned long long int num   285212672    FYI  fits in 29 bits     int normalInt   5      printf  My number is  d bytes wide and its value is  ul  A normal number is  d  n   sizeof num   num  normalInt       return 0      Output   My number is 8 bytes wide and its value is 285212672l  A normal number is 0    I assume this unexpected result is from printing the unsigned long long int  How do you printf   an unsigned long long int

User · Answer

Apparently no one has come up with a multi-platform* solution for over a decade since [the] year 2008, so I shall append mine . Plz upvote. (Joking. I don’t care.)

Solution: `lltoa()`

How to use:

#include <stdlib.h> /* lltoa() */
// ...
char dummy[255];
printf("Over 4 bytes: %s\n", lltoa(5555555555, dummy, 10));
printf("Another one: %s\n", lltoa(15555555555, dummy, 10));

OP’s example:

#include <stdio.h>
#include <stdlib.h> /* lltoa() */

int main() {
    unsigned long long int num = 285212672; // fits in 29 bits
    char dummy[255];
    int normalInt = 5;
    printf("My number is %d bytes wide and its value is %s. "
        "A normal number is %d.\n", 
        sizeof(num), lltoa(num, dummy, 10), normalInt);
    return 0;
}

Unlike the %lld print format string, this one works for me under 32-bit GCC on Windows.

*) Well, almost multi-platform. In MSVC, you apparently need _ui64toa() instead of lltoa().

User · Answer

For long long  or   int64  using MSVS  you should use  I64d     int64 a  time t b      fprintf outFile   I64d  I64d n  a b        I is capital i

User · Answer

Non-standard things are always strange     for the long long portion under GNU it s L  ll or q  and under windows I believe it s ll only

User · Answer

That is because  llu doesn t work properly under Windows and  d can t handle 64 bit integers  I suggest using PRIu64 instead and you ll find it s portable to Linux as well   Try this instead    include  lt stdio h gt   include  lt inttypes h gt   int main         unsigned long long int num   285212672    FYI  fits in 29 bits     int normalInt   5         NOTE  PRIu64 is a preprocessor macro and thus should go outside the quoted string         printf  My number is  d bytes wide and its value is    PRIu64    A normal number is  d  n   sizeof num   num  normalInt       return 0      Output  My number is 8 bytes wide and its value is 285212672  A normal number is 5

User · Answer

How do you format an unsigned long long int using printf   Since C99 use an  quot ll quot   ell-ell  before the conversion specifiers o u x X  In addition to base 10 options in many answers  there are base 16 and base 8 options  Choices include unsigned long long num   285212672  printf  quot Base 10   llu n quot   num   num    0xFFF     For more interesting hex octal output  printf  quot Base 16   llX n quot   num      Use uppercase A-F printf  quot Base 16   llx n quot   num      Use lowercase a-f printf  quot Base  8   llo n quot   num   puts  quot or 0x 0X prefix quot    printf  quot Base 16    llX   llX n quot   num  0ull      When non-zero  print leading 0x printf  quot Base 16    llx   llx n quot   num  0ull   printf  quot Base 16  0x llX n quot   num      My hex fave  lower case prefix  with A-F  Output Base 10  285212672 Base 16  11000FFF Base 16  11000fff Base  8  2100007777 or 0x 0X prefix Base 16  0X11000FFF 0 Base 16  0x11000fff 0 Base 16  0x11000FFF

User · Answer

In addition to what people wrote years ago    you might get this error on gcc mingw    main c 30 3  warning  unknown conversion type character  l  in format  -Wformat    printf   llu n   k    Then your version of mingw does not default to c99  Add this compiler flag  -std c99

User · Answer

Well  one way is to compile it as x64 with VS2008  This runs as you would expect   int normalInt   5   unsigned long long int num 285212672  printf       My number is  d bytes wide and its value is  ul       A normal number is  d  n        sizeof num        num       normalInt     For 32 bit code  we need to use the correct   int64 format specifier  I64u  So it becomes    int normalInt   5   unsigned   int64 num 285212672  printf       My number is  d bytes wide and its value is  I64u       A normal number is  d        sizeof num       num  normalInt     This code works for both 32 and 64 bit VS compiler

User · Answer

d--  for int   u--  for unsigned int   ld--  for long int or long   lu--  for unsigned long int or long unsigned int or unsigned long   lld--  for long long int or long long   llu--  for unsigned long long int or unsigned long long

User · Answer

Compile it as x64 with VS2005       llu works well

User · Answer

You may want to try using the inttypes h library that gives you types such as int32 t  int64 t  uint64 t etc  You can then use its macros such as   uint64 t x  uint32 t y   printf  x    PRId64   y    PRId32  n   x  y     This is  guaranteed  to not give you the same trouble as long  unsigned long long etc  since you don t have to guess how many bits are in each data type

User · Answer

Hex   printf  64bit   llp   0xffffffffffffffff     Output   64bit  FFFFFFFFFFFFFFFF

User · Answer

Use the ll  el-el  long-long modifier with the u  unsigned  conversion   Works in windows  GNU    printf   llu   285212672

User · Answer

In Linux it is  llu and in Windows it is  I64u  Although I have found it doesn t work in Windows 2000  there seems to be a bug there

[c] How do you format an unsigned long long int using printf?

Solution: `lltoa()`

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[c] How do you format an unsigned long long int using printf?

Solution: lltoa()

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Solution: `lltoa()`