C //例3.2 计算存款利息。有1000元，想存一年。有3种方法可选：。。。

可选：。。。"/>

C //例3.2 计算存款利息。有1000元，想存一年。有3种方法可选：。。。

C程序设计（第四版） 谭浩强 例3.2

例3.2 计算存款利息。有1000元，想存一年。有3种方法可选：

（1）活期，年利率为r1；

（2）一年期定期，年利率为r2；

（3）存两次半年定期，年利率为r3。

请分别计算出一年后按3种方法所得到的本息和。

IDE工具：VS2010

Note: 使用不同的IDE工具可能有部分差异。

代码块

方法1：

#include <stdio.h>
#include <stdlib.h>int main(){double *compoundInterest = (double*)malloc(3 * sizeof(double));double principal = 1000.0;double r1 = 0.0036;double r2 = 0.0225;double r3 = 0.0198;compoundInterest[0] = principal * (1 + r1);compoundInterest[1] = principal * (1 + r2);compoundInterest[2] = principal * (1 + r3 / 2) * (1 + r3 / 2);for(int i = 0; i < 3; i++){printf("Compound Interest %d = %.2f\n", i + 1, compoundInterest[i]);}free(compoundInterest);system("pause");return 0;
}

方法2：使用函数指针、函数的模块化设计

#include <stdio.h>
#include <stdlib.h>double fun1(double principal, double r1){return principal * (1 + r1);
}double fun2(double principal, double r2){return principal * (1 + r2);
}double fun3(double principal, double r3){return principal * (1 + r3 / 2) * (1 + r3 / 2);
}void compoundInterest(double principal, double *r, double (*fun[])(double, double)){for(int i = 0; i < 3; i++){printf("Compound Interest %d = %.2f\n", i + 1, fun[i](principal, r[i]));}
}int main(){double (*fun[3])(double, double) = {fun1, fun2, fun3};double principal = 1000.0;double *r = (double*)malloc(3 * sizeof(double));r[0] = 0.0036;r[1] = 0.0225;r[2] = 0.0198;compoundInterest(principal, r, fun);free(r);system("pause");return 0;
}