税级计算器自我分配

Question

我的任务是以任何可能的方式创建一个税收计算器，最初是在不包括税级的情况下纳税。我将如何调整此代码以改用循环或如何使其更高效？

#include <stdio.h>
#include <stdlib.h>

int main()
{
    double earnings;    //user income
    double tax;         //sum of total tax

    printf("=====================<><>TAX CALCULATOR (INC TAX BRACKETS)<><>=====================\n How much have you earned: \n");
    scanf("%lf", &earnings);
    double tax1 = earnings -12000;      //seperating each tax bracket
    double tax2 = earnings - 20000;
    double tax3 = earnings - 40000;
    double tax4;
    double tax5;
    if (earnings >= 12000 && earnings <=19999){
        tax = (tax1*0.1);
        printf("\nYou've made: %0.2f", earnings - tax);
        printf("\nPaying: %0.2f in tax", tax);
    }
    else if(earnings >= 20000 && earnings <=39999){
        //printf("\n============tax2: %0.2f============\n",tax2);
        //printf("\n============Tax2: %0.2f============\n",tax2*0.15);
        tax = (tax1*0.1 + tax2*0.15);
        printf("\nYou've made: %0.2f", earnings - tax);
        printf("\nPaying: %0.2f in tax", tax);
    }
    else if(earnings >= 40000){
        tax = (tax1*0.1 + tax2*0.15 + tax3*0.2);
        //printf("\n============tax3: %0.2f============\n",tax3);
        //printf("\n============Tax3: %0.2f============\n",tax3*0.2);
        printf("\nYou've made: %0.2f", earnings - tax);
        printf("\nPaying: %0.2f in tax", tax);
    }
    else{
            tax = 0;
            printf("You made %0.2f", earnings);
            printf("Paying: %0.2f", tax);
    }

    return 0;
}

Answer 1

您可以简化您的条件，例如：

    if (earnings < 12000) {
    } else if (earnings < 20000) {
    } else if (earnings < 40000) {
    } else { // earnings >= 40000 
    }

您还可以减少“有时不需要”的预计算，合并对 printf 的连续调用并避免在任何地方重复它们，并将运行-time 计算替换为编译时计算，如：

    const double constTax1 = (20000 - 12000) * 0.1;
    const double constTax2 = (40000 - 20000) * 0.25 + constTax1;
    double tax = 0.0;

    if (earnings < 12000) {
        // Do nothing!
    } else if (earnings < 20000) {
        tax = (earnings - 12000) * 0.1;
    } else if (earnings < 40000) {
        tax = (earnings - 20000) * 0.25 + constTax1;
    } else { // earnings >= 40000 
        tax = (earnings - 20000) * 0.45 + constTax2;
    }
    printf("\nYou've made: %0.2f\nPaying: %0.2f in tax", earnings - tax, tax);

幸运的是，编译器很可能会自己完成大部分优化。最大的性能提升将来自合并对 printf() 的调用（对 reduce/avoid 函数调用开销）。

请注意，编译器通常无法正确优化对 printf() 的调用，因为它位于不同的编译单元中（即使您使用 link 时间优化，也可能 printf()在动态 linked 库中。

Answer 2

您可以执行类似以下操作，以避免多次写入相同的数字并更容易添加额外的税级。（这几乎肯定不会让它变得更快，但我确实认为它可能更易于维护。）

const int nb = 3;
const double taxrate[nb] = {0.1, 0.15, 0.2};
const double taxcutoff[nb] = {12000, 20000, 40000};
double tax = 0;

for (int i = 0; i < nb && earnings > taxcutoff[i]; i++) {
 tax += (earnings - taxcutoff[i]) * taxrate[i];
}
printf("\nYou've made: %0.2f", earnings - tax);
printf("\nPaying: %0.2f in tax", tax);

我相信以上内容与您的原始代码相同，但这并不是真正的税级的工作方式。您更有可能想要这样的东西：

const int nb = 3;
const double taxrate[nb] = {0.1, 0.15, 0.2};
const double taxcutoff[nb] = {12000, 20000, 40000};
double bracketearnings;
double tax = 0;

for (int i = 0; i < nb && earnings > taxcutoff[i]; i++) {
 if (i == nb-1 || earnings < taxcutoff[i+1]) {
    bracketearnings = earnings - taxcutoff[i];
 }
 else {
    bracketearnings = taxcutoff[i+1] - taxcutoff[i];
 }
 tax += bracketearnings * taxrate[i];
}
printf("\nYou've made: %0.2f", earnings - tax);
printf("\nPaying: %0.2f in tax", tax);

编辑以按要求解释循环：在这两种情况下，我都使用 nb=3 来跟踪有多少税级。循环将遍历数组的每个项目（或直到“收入 > taxcutoff[I]”不成立）。

在第二个示例中，它检查下一个税级是否不存在 (i == nb-1) 或收入是否“结束”在此税级中 (earnings < taxcutoff[i+1])。如果是这样，它使用收入减去当前税收截止作为 bracketearnings，否则它使用下一个括号截止减去当前截止作为 bracketearnings。在任何一种情况下，它都会将 bracketearnings 乘以当前税率并将其添加到 tax。税收是一个运行总额，在循环的每次迭代中都会添加。

与您最初的想法相比的优势在于，如果您想添加另一个税级（或 100 个以上），您只需更改前 3 行，因为它们每一行的逻辑都是相同的。

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