扫描和打印 uint64_t (long long unsigned int) 时遇到问题

Question

#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>
#include <stdlib.h>

int main(void)
{
    uint64_t *ptr = malloc(sizeof(uint64_t));
    scanf("%llu",&ptr);
    printf("%llu\n", *ptr);
    free(ptr);
}

The compiler says that

mod_5_working.c:9:14: error: unknown conversion type character 'l' in format [-Werror=format=]
     scanf("%llu",&ptr);
              ^
mod_5_working.c:9:11: error: too many arguments for format [-Werror=format-extra-args]
     scanf("%llu",&ptr);

I've tried using %u but it says that I should use %llu .

Answer 1

• scanf("%llu",&ptr); is senseless, like the compiler tells you, you are taking the address of a pointer.

• uint64_t doesn't necessarily correspond to unsigned long long . Some 64 bit systems use unsigned long for 64 bit numbers.

The correct, portable specifiers to use when scanning/printing uint64_t is SCNu64 and PRIu64 from inttypes.h:

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <stdlib.h>

int main(void)
{
    uint64_t *ptr = malloc(sizeof(uint64_t));
    scanf("%"SCNu64, ptr);
    printf("%"PRIu64"\n", *ptr);
    free(ptr);
}

Answer 2

scanf expects a pointer, not a pointer to a pointer. Simply pass the pointer to scanf , no need to & it. Also you should always check the return value of malloc for errors.

uint64_t *ptr = malloc(sizeof(uint64_t));
if (!ptr)
    return -1; /* Maybe handle it better */

scanf("%" SCNu64, ptr);
printf("%" PRIu64 "\n", *ptr);

free(ptr);

Edit: It's a better practice to use SCNu64 and PRIu64 (from <inttypes.h> ) to make it portable and accurate.

Answer 3

Use the matching specifier.

#include <stdio.h>
#include <inttypes.h> /* Format conversions for exact-width types */

scanf("%" SCNu64, ptr);  // No &.
printf("%" PRNu64 "\n", *ptr);