c - Hogyan formázhatja az alá nem írt hosszú, hosszú int-et a printf segítségével?

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

printf("%llu", 285212672);

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().