How do you format an unsigned long long int using printf?
-
08-06-2019 - |
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
?
Solution
Use the ll (el-el) long-long modifier with the u (unsigned) conversion. (Works in windows, GNU).
printf("%llu", 285212672);
OTHER TIPS
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.
%d
--> for int
%u
--> for unsigned int
%ld
--> for long int
%lu
--> for unsigned long int
%lld
--> for long long int
%llu
--> for unsigned long long int
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
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.
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!
Compile it as x64 with VS2005:
%llu works well.
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
Hex:
printf("64bit: %llp", 0xffffffffffffffff);
Output:
64bit: FFFFFFFFFFFFFFFF
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
.
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.
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()
.