如何在uint64_t上使用算术运算？

Question

我正在尝试制作一个程序，该程序将给定数字简单地除以并打印出余数，并将给定数字的解决方案除以10。但是我的代码并没有打印出正确的值。 这是下面的代码：

#include <stdio.h>
#include <string.h>
#include <stdint.h>
#include <stdlib.h>
uint64_t divide(uint64_t,uint64_t);

int main(int argc, char *argv[])
{

        uint64_t num1 = 224262;
        uint64_t num2 = 244212;
        divide(num1,num2);

}


uint64_t divide( uint64_t set1, uint64_t set2 )
{

        printf("%lX\n",set1);
        uint64_t remainder = set1%10;
        printf("%lX\n",remainder);
        set1= set1/10;
        printf("%lX\n",set1);
}

目前，此输出给我以下内容

36C06
2
579A

我将如何获得它以便正确输出除数值和余数？

Answer 1

假设您的平台具有64 bits long ，则printf的正确格式为%lu

uint64_t divide( uint64_t set1, uint64_t set2 )
{
    printf("%lu\n",set1);
    uint64_t remainder = set1%10;
    printf("%lu\n",remainder);
    set1= set1/10;
    printf("%lu\n",set1);

    return set1;
}

使用inttypes.h预定义格式说明符，您可以编写

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

uint64_t divide(uint64_t,uint64_t);

int main(int argc, char *argv[])
{
    uint64_t num1 = 224262;
    uint64_t num2 = 244212;

    divide(num1,num2);
}


uint64_t divide( uint64_t set1, uint64_t set2 )
{
    printf("%"PRIu64"\n",set1);
    uint64_t remainder = set1%10;
    printf("Reminder = %"PRIu64"\n",remainder);
    set1= set1/10;
    printf("Division = %"PRIu64"\n",set1);

    return set1;
}