Question

#include <stdio.h>


void temperature(double x) {
    double f = ((9.0/5)*(x))+32;
    return f;
}

int main (int argc , char * argv[]){
    printf("Please Enter a degree in Celsius =>");

    if (argc > 1){
        double c = atolf(argv[1]);

        double result = temperature(c);

        printf("%lf celcius in fahrenheit is %lf",c,result);

        }
    else {
        printf("Please enter a temperature in degrees celius");
    }

}

错误（在cygwin中编译时）我认为问题出在void方法温度：我试图在温度方法中返回一个double值。谢谢你的帮助。

$ gcc -o temperature temperature.c
temperature.c: In function 'temperature':
temperature.c:6: warning: 'return' with a value, in function returning void
temperature.c: In function 'main':
temperature.c:15: error: void value not ignored as it ought to be

Answer 1

由于函数返回double，您需要相应地更改其返回类型：

double temperature(double x) {
^^^^^^

使用void方法有小问题

1 个答案: