以下代碼有什麼問題？

Question

//，經過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

發生很大的變化：

scanf("%d", &money1); 

到

scanf("%lf", &money1); 

和變化：

scanf("%d", &apy1); 

到：

scanf("%lf", &apy1); 

雖然你在它你可能想添加一些printfs輸出，以幫助調試（假設你沒有源代碼級調試。 ）

Answer 2

這在計算銀行賬戶的金額的程序：

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

也許應該

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

，因爲你賺利息應該添加到現有的金額。否則，您的總金額將每年減少。

Answer 3

我想你應該叫你按以下方式運作：

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

否則你有很多錢:)

Answer 4

另一條是：

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

返回double，但

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

將結果放在int中。

Answer 5

You should never use floating point in financial calculations.

浮點固有uncapable精確地表示10個鹼基的值，這意味着你將舍入誤差和unequalities，這是不可接受的財政（等等）受到影響。

This has been discussed in detail many times on SO。這個問題不是特定於語言的。