在 C 中将函数作为 float 调用

Question

我正在使用 C 中的函数。如果我想返回 void taxCalculator();，代码会是什么样子？将 int main() 转换为 float ，以便可以在那里打印。

代码是这样的:

顶部的定义

void taxCalculator(float income, float tax);

主要功能:

int main(void){
float income, tax;

printf("Enter the amount of income: ");    
scanf("%f", &income);
taxCalculator(income, tax);
}

还有 taxCalculator 函数:

void taxCalculator(float income, float tax)
{
if( income < 750.00f){
    tax = income * 0.01f;
}else if (income <= 2250.00f){
    tax = 7.50f + (income - 750) * 0.02f;
}else if (income <= 3750.00f){
    tax = 37.50f + (income - 2250) * 0.03f;
}else if (income <= 5250){
    tax = 82.50f + (income - 3750) * 0.04f;
}else if (income <= 7000){
    tax = 142.50f + (income - 5250) * 0.05f;
}else if(income > 7000.00f){
    tax = 230.00f + (income - 7000) * 0.06f;
}
printf("The amount of tax: %.2f", tax);

}

Answer 1

你可以这样做

float taxCalculator(float income) {
  ...
  return tax;
}

...

printf("The amount of tax: %.2f", taxCalculator(income));

当函数执行时，当它终止时，它将被替换为它的返回值，因此 printf() 将使用该值进行打印。

---

完整示例:

#include <stdio.h>

float taxCalculator(float income) {
  float tax;
  if (income < 750.00f) {
    tax = income * 0.01f;
  } else if (income <= 2250.00f) {
    tax = 7.50f + (income - 750) * 0.02f;
  } else if (income <= 3750.00f) {
    tax = 37.50f + (income - 2250) * 0.03f;
  } else if (income <= 5250) {
    tax = 82.50f + (income - 3750) * 0.04f;
  } else if (income <= 7000) {
    tax = 142.50f + (income - 5250) * 0.05f;
  } else if (income > 7000.00f) {
    tax = 230.00f + (income - 7000) * 0.06f;
  }
  return tax;
}

int main(void) {
  float income = 2250;
  printf("The amount of tax: %.2f", taxCalculator(income));
  return 0;
}