How do you printf an unsigned long long int?

Question

#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?

Answer 1

Use the ll (el-el) long-long modifier with the u (unsigned) conversion.

printf("%llu", 285212672);

Answer 2

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

Answer 3

%llu does not work either. Note that normalInt outputs as 0, instead of 5 as it should.

#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 %llu. 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 0.

Answer 4

%llu does not work either. Note that normalInt outputs as 0, instead of 5 as it should

@superjoe30: This looks like the %llu is working and printing your unsigned long long int just fine.

it's the %d => normal int which isn't working, which (although I thought %d handled ints just fine, this may change with different compilers etc) you might just need to change to %i

Check the manual for printf for your compiler

Answer 5

@Orion Edwards:

%d is the same as %i, but just to make sure I tried compiling it, and yes, it gives the same results. It is, in fact, the int64 causing problems.

@vzczc:

But I don't want to compile it with 64-bit. It's a 32-bit application.

FYI The code that I have posted here was the output of using gcc.

Answer 6

std::cout might work for this. (I'm at work and cant test it)

Answer 7

I just compiled your code ( with %llu ) with gcc and the output was the correct one.

Are you passing any options to the compiler?

Answer 8

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.

Answer 9

@superjoe30: What platform are you on? On ubuntu gutsy, your program (with %llu) gives the following output:

$ gcc -o num num.c
$ ./num 
My number is 8 bytes wide and its value is 285212672. A normal number is 5.

I also get the same output on debian etch. Both machines are 32 bit intels.

Answer 10

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;
    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.

Answer 11

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!

Answer 12

on HPUX machine, I have some other problem.

#include<stdio.h>
#include <inttypes.h>
void main()
{
        /*unsigned long var;*/
        unsigned long long var;
        var = 4294967295;
        printf("Sizeof uint = %d\n",sizeof(var));
        printf("var = %"PRIu64"\n",var);
}

output:

#cc one.c
# ./a.out
Sizeof uint = 8
var = 18446744073709551615

how is this possible, i have assigned a value to "var" why is it not getting printed?? Please help me.

Answer 13

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

Answer 14

correct your line to: printf("var = %PRIu64\n",var);

Answer 15

Note that samsung bada's newlib seems not to support "%lld" :

http://developer.bada.com/forum/topic.php?id=2966&replies=7#post-22945