scanf后的SegFault? [英] SegFault after scanf?

问题描述

#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之后会出现分段错误?

Why does this have a segmentation fault when after scanf?

推荐答案

scanf 需要将变量的地址传递给它.用

scanf requires the addresses of the variables to be passed to it. Replace your scanf by

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

你应该很高兴.

这篇关于scanf后的SegFault?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！