以下代码有什么问题？

Question

//，经过N年

#include <stdio.h> 

double bank(double money, double apy, int years); 

int main() { 

double money1, apy1; 
int years1; 

printf("How much money is currently in your bank account? "); 
scanf("%d", &money1); 

printf("How many years will this money stay in your account? "); 
scanf("%d",&years1); 

printf("What is your APY? "); 
scanf("%d", &apy1); 

int bank1 = bank(money1, apy1, years1); 

printf("Your grand total after %d will be $%d \n", years1, bank1); 


system ("PAUSE"); 
return 0; 
} 


double bank(double money, double apy, int years) { 

if(years <= 0) 
    return money; 

else 
    return bank(money*apy, apy, years-1); 

} 

Answer 1

发生很大的变化：

scanf("%d", &money1); 

到

scanf("%lf", &money1); 

和变化：

scanf("%d", &apy1); 

到：

scanf("%lf", &apy1); 

虽然你在它你可能想添加一些printfs输出，以帮助调试（假设你没有源代码级调试。 ）

Answer 2

这在计算银行账户的金额的程序：

return bank(money*apy, apy, years-1); 

也许应该

return bank(money*(1+apy), apy, years-1); 

，因为你赚利息应该添加到现有的金额。否则，您的总金额将每年减少。

Answer 3

我想你应该叫你按以下方式运作：

int bank1 = bank(money1, 1+apy1/100., years1); 

否则你有很多钱:)

Answer 4

另一条是：

double bank(double money, double apy, int years); 

返回double，但

int bank1 = bank(money1, apy1, years1); 

将结果放在int中。

Answer 5

You should never use floating point in financial calculations.

浮点固有uncapable精确地表示10个碱基的值，这意味着你将舍入误差和unequalities，这是不可接受的财政（等等）受到影响。

This has been discussed in detail many times on SO。这个问题不是特定于语言的。