扫描后的SegFault？

Question

#include <stdio.h>
#define TimeConverter 60
#define TempFormula time * time * 4 / time + 2 - 20

double HoursMinToTime(int hour, int min);

double Temperature(double time);

int main()
{
    int hour, min;
    double time, temperature;
    printf("Hours and minutes: ");
    scanf("%d %d", hour, min);
    //Segfault HERE
    time = HoursMinToTime(hour, min);
    temperature = Temperature(time);
    printf("After a %lf hour power failure, the frezer will be %lf degrees", time, temperature);
    return 0;
}
double HoursMinToTime(int hour, int min)
{
    double time = hour * min / TimeConverter;
    return time;              
}
double Temperature(double time)
{
    double temp;
    temp = TempFormula;
    return temp;
}

为什么在scanf之后会有一个分段错误？

Answer 1

scanf需要将变量的地址传递给它。将scanf替换为

scanf("%d %d",&hour,&min);

你应该可以走了。

Answer 2

您需要传递这些变量的地址。一定要特别注意函数签名，因为你可能会得到非常奇怪的结果。另外，打开编译器的所有警告。

scanf("%d %d", &hour, &min); 

Answer 3

scanf("%d %d", &hour, &min);