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

问题

#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?

回答1:

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

printf("%llu", 285212672);

回答2:

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.

回答3:

%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

回答4:

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

回答5:

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.

回答6:

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!

回答7:

Compile it as x64 with VS2005:

%llu works well.

回答8:

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

回答9:

Hex:

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

Output:

64bit: FFFFFFFFFFFFFFFF

回答10:

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.

回答11:

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.

回答12:

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().

来源：https://stackoverflow.com/questions/2844/how-do-you-format-an-unsigned-long-long-int-using-printf

标签

syntax

printf

format-specifiers

long-long