Question

我写的很快，每次执行程序时，我最终都会得到32华氏度和0.00摄氏度，我不知道这里有什么问题。

 #include <stdio.h>
 int main(void)
 {
     double celsius=0, fahrenheit=0;
     printf("Enter a temperature in degrees Celsius: ");
     scanf("%f", &celsius);
     fahrenheit =(5.0/9.0)*celsius + 32;
     printf("That is %.2f Fahrenheit \n", fahrenheit);
     printf("Enter a temperature in degrees fahrenheit: ");
     scanf("%f", &fahrenheit);
     celsius = (fahrenheit - 32) * (5.0/9.0);
     printf("That is %.2f Celsius \n", celsius);
     return(0);
 }

Answer 1

double的格式说明符错误，应为"%lf"。但这还不够，你还应该检查scanf()成功读取读取值，就像这样

#include <stdio.h>

int report_error(const char *const message)
{
    // TODO: add message formatting capabilities to this function
    fprintf(stderr, "error: %s\n", message);
    return EXIT_FAILURE;
}

int main(void)
{
    double celsius = 0;
    double fahrenheit = 0;

    printf("Enter a temperature in degrees Celsius: ");
    if (scanf("%lf", &celsius) != 1)
        return report_error("Invalid Input");
    fahrenheit = (5.0 / 9.0) * celsius + 32;
    //                ^ Define this as a constant?

    printf("That is %.2f Fahrenheit \n", fahrenheit);
    printf("Enter a temperature in degrees fahrenheit: ");

    if (scanf("%lf", &fahrenheit) != 1)
        return report_error("Invalid Input");
    celsius = (fahrenheit - 32) * (5.0 / 9.0);
    //                                 ^ see if it was a constant???

    printf("That is %.2f Celsius \n", celsius);
    return 0;
}

此外，您的代数似乎出错了。您应该明确检查celsius表达式。

C转换为Farhenheit转换为摄氏度

1 个答案: