Question

我似乎无法绕过这个。我有其他程序的相同设置，这是一个堆栈，推/弹，它的工作完美。我在函数错误中首次使用未声明的值。任何帮助，将不胜感激。

头文件

#include <stdio.h>
#include <string.h>

int money();
void amortization();

typedef struct{
int principle;
int rate;
int payments;
} loan_t;

功能代码

int money(loan_t)
{
 printf("Please input the amount borrowed:");
 scanf("%d", &principle);
 printf("\nPlease input the Annual Interest Rate:");
 scanf("%d", &rate);
 printf("\nPlease input the number of monthly payments:\n");
 scanf("%d", &payments);
 return (principle,rate,payments);
 }

谢谢你！

Answer 1

尝试：

 int money(LOAN *var){
     printf("Please input the amount borrowed:");
     scanf("%lf", var->principle);
     printf("\nPlease input the Annual Interest Rate:");
     scanf("%lf", var->rate);
     printf("\nPlease input the number of monthly payments:\n");
     scanf("%lf", var->payments);
     return 0;
 }

将标题更改为：

#include <stdio.h>
#include <string.h>

typedef struct loan{
    double principle;
    double rate;
    double payments;
}LOAN;

int money(LOAN *var);
void amortization();

这意味着您将在

中返回具有所需值的loan_t

编辑：编辑以最适合您，而不是Jonathan Leffler推荐的最佳实践

在LOAN variable;内声明main()，然后致电money(&variable);

Answer 2

根据您提供的内容，principle中的rate，payments和money似乎都未声明。您还在money中提供了一个没有参数名称的参数作为参数。

编辑：正如另一位回答者所指出的那样，回归也很可疑。

我现在无法自己测试，但请尝试：

loan_t money(loan_t loan)
{
 printf("Please input the amount borrowed:");
 scanf("%lf", &(loan.principle));
 printf("\nPlease input the Annual Interest Rate:");
 scanf("%lf", &(loan.rate));
 printf("\nPlease input the number of monthly payments:\n");
 scanf("%lf", &(loan.payments));
 return loan;
 }

头文件创建未声明的值

2 个答案: