How do you format an unsigned long long int using printf?

395

#include <stdio.h>
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?

This question is tagged with c syntax printf format-specifiers long-long

~ Asked on 2008-08-05 20:59:29

13 Answers


505

Use the ll (el-el) long-long modifier with the u (unsigned) conversion. (Works in windows, GNU).

printf("%llu", 285212672);

~ Answered on 2008-08-05 21:02:35


94

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.

~ Answered on 2008-08-12 11:07:50


88

%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

~ Answered on 2015-05-13 17:48:33


40

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

~ Answered on 2009-09-06 14:26:41


37

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 <stdio.h>
#include <inttypes.h>

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.

~ Answered on 2008-09-09 07:54:37


15

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!

~ Answered on 2008-09-11 00:57:55


8

Compile it as x64 with VS2005:

%llu works well.

~ Answered on 2012-10-11 06:23:13


3

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().

~ Answered on 2019-03-22 19:08:07


3

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.

~ Answered on 2016-01-17 12:27:29


1

How do you format an unsigned long long int using printf?

Since C99 use an "ll" (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("Base 10: %llu\n", num);
num += 0xFFF; // For more interesting hex/octal output.
printf("Base 16: %llX\n", num); // Use uppercase A-F
printf("Base 16: %llx\n", num); // Use lowercase a-f
printf("Base  8: %llo\n", num);
puts("or 0x,0X prefix");
printf("Base 16: %#llX %#llX\n", num, 0ull); // When non-zero, print leading 0x
printf("Base 16: %#llx %#llx\n", num, 0ull);
printf("Base 16: 0x%llX\n", 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

~ Answered on 2021-02-03 13:33:30


1

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

~ Answered on 2008-08-05 21:03:07


0

Hex:

printf("64bit: %llp", 0xffffffffffffffff);

Output:

64bit: FFFFFFFFFFFFFFFF

~ Answered on 2015-07-26 09:57:06


0

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.

~ Answered on 2008-08-05 21:40:24


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