scanf函数后段错误? [英] SegFault after scanf?
本文介绍了scanf函数后段错误?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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;
}
为什么这个scanf函数之后,当有一个分段错误?
Why does this have a segmentation fault when after scanf?
推荐答案
的 scanf函数的要求变量的地址传递给它。通过
scanf requires the addresses of the variables to be passed to it. Replace your scanf by
scanf("%d %d",&hour,&min);
您应该是好去。
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