c - 在华氏、摄氏和开尔文之间转换的程序

Question

我正在寻找有关 C 程序的帮助。我们的教授向我们展示了一个示例，我们将输入摄氏度或华氏度的温度并将其转换为其他温度。我发现它很有趣，并尝试更进一步，添加开尔文。

#include <stdio.h>
int main(void)
{
    #define MAXCOUNT 4
    float tempConvert(float, char);

    int count;
    char my_char;
    float convert_temp, temp;
    for(count = 1; count <= MAXCOUNT; count++)
    {
        printf("\nEnter a temperature: ");
        scanf("%f %c", &temp, &my_char);
        convert_temp = tempConvert(temp, my_char);
        if (my_char == 'c')
            printf("The Fahrenheit equivalent is %5.2f degrees\n"
                "The Kelvin equivalent is %5.2f degrees\n",
                convert_temp,convert_temp);
        else if (my_char == 'f')
            printf("The Celsius equivalent is %5.2f degrees\n"
                "The Kelvin equivalent is %5.2f degrees\n",
                convert_temp,convert_temp);
        else if (my_char == 'k')
            printf("The The Celsius equivalent is %5.2f degrees\n"
                "The Fahrenheit equivalent is %5.2f degrees\n",
                convert_temp,convert_temp);
    }
    return 0;
}

float tempConvert(float inTemp, char ch)
{
    float c_temp1, c_temp2;
    if (ch == 'c'){
        return c_temp1 = ( (5.0/9.0) * (inTemp - 32.0) );
        return c_temp2 = ( inTemp + 273.15 );}
    else if (ch == 'f'){
        return c_temp1 = ( ((9.0/5.0) * inTemp ) + 32.0 );
        return c_temp2 = ( (5.0/9.0) * (inTemp + 459.67 ) );}
    else if (ch == 'k'){
        return c_temp1 = ( inTemp - 273.15 );
        return c_temp2 = ( ((9.0/5.0) * inTemp ) - 459.67 );}
}

该程序正在终端中运行，但问题是我只得到第一次温度转换的答案，而不是第二次的答案(第二次与第一次相同)。我的问题是为什么第二个答案没有被识别，我该如何解决？

Answer 1

您在一个分支中返回不止一次。因此，只有第一个 return 语句被执行。

您不能从一个函数中返回几个整数。但是您可以分配一个输出参数数组。我会这样做(为什么不用 switch/case 语句？):

void tempConvert(float inTemp, char ch, float results[2])
{
    switch(ch)
    {
    case 'c':
        results[0] = ( (5.0/9.0) * (inTemp - 32.0) );
        results[1]  = ( inTemp + 273.15 );
        break;
    case 'f':
        results[0] = ( ((9.0/5.0) * inTemp ) + 32.0 );
        results[1]  = ( (5.0/9.0) * (inTemp + 459.67 ) );
        break;
    case 'k':
         results[0]  = ( inTemp - 273.15 );
         results[1]  = ( ((9.0/5.0) * inTemp ) - 459.67 );
         break;
    default:
         results[0] = results[1] = 0; // kind of error code
  }
}

这样调用它:

float convert_temp[2]
tempConvert(temp, my_char, convert_temp);

if (my_char == 'c')
    printf("The Fahrenheit equivalent is %5.2f degrees\n"
        "The Kelvin equivalent is %5.2f degrees\n",
        convert_temp[0],convert_temp[1]);