如何在const char字符串中转换uint64_t值？

Question

See in one situation

uint64_t trackuid = 2906622092;

Now I want to pass this value in one function where function argument is const char*

func(const char *uid)
{
   printf("uid is %s",uid);
}

This should print

uid is 2906622092 

How can I do this?

Answer 1

// length of 2**64 - 1, +1 for nul.
char buff[21];

// copy to buffer
sprintf(buff, "%" PRIu64, trackuid);

// call function
func(buff);

This requires C99, however, my memory says the MS compiler doesn't have PRIu64 . ( PRIu64 is in inttypes.h .) YMMV.

Answer 2

Use snprintf to convert numbers to strings. For integer types from stdint.h header use the format macros from inttypes.h .

#define __STDC_FORMAT_MACROS // non needed in C, only in C++
#include <inttypes.h>
#include <stdio.h>

void func(const char *uid)
{
    printf("uid is %s\n",uid);
}

int main()
{
    uint64_t trackuid = 2906622092;

    char buf[256];
    snprintf(buf, sizeof buf, "%"PRIu64, trackuid);

    func(buf);

    return 0;
}

Answer 3

char buf[40];
memset (buf, 0, sizeof(buf));
snprintf (buf, sizeof(buf)-1, "%llu", (unsigned long long) trackuid);
func(buf);

should work when sizeof(unsigned long long) == sizeof(uint64_t)

EDIT

but the better answer is by Maxim Yegorushkin to use "%"PRIu64