c - c中的开关函数

Question

尝试创建一个菜单驱动的员工数据程序。我不知道如何创建一个功能菜单，并且也无法使菜单选项正常工作，例如编辑之前输入的员工信息。如果我能在这方面得到任何帮助，我将不胜感激。目前出现的错误是

1. 编辑员工功能:重新定义；不同的基本类型。

2. 在main中——调用我正在使用menu(&payroll)的菜单函数时，错误消息是，无法从input*<转换 到输入。

3. 同样在 main 中，函数 employeeInfo(&payroll) 给出错误消息，无法从 input* 转换为 input .

我确信我还犯了很多其他错误，如果您发现任何错误，请引导我走向正确的方向。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//This is a macro intended for use with the emplyName array.
#define SIZE 20 

//This struct has all the varibles that I will be using in my functions
typedef struct
{
    char emplyName[5][SIZE];
    float emplyHours[5];
    float emplyRate[5];
    float emplyGross[5];
    float emplyBase[5];
    float emplyOvrt[5];
    float emplyTax[5];
    float emplyNet[5];
    float emplyTotal[5];
}input;

void menu(void);
void employeeInfo(input* emply);
void editEmployees(input* emply);
void print(input* emply);

int main(void)
{
    input payroll={"",0.0f,0.0f,0.0f,0.0f,0.0f,0.0f,0.0f,0.0f};
    int choice;
    menu(&payroll);
    scanf_s("%d", &choice);
    switch (choice){
    case '1': 
        employeeInfo(&payroll);
        break;
    case '2':
        editEmployees(&payroll);
        break;
    case '3':
        break;
    case '4':
        print(&payroll);
        break;
    default:
        printf("Invalid entry\n");
    }
    system("pause");
}

void employeeInfo(input *emply)
{
    int i;

    for (i = 0; i < 5; i++){
        printf("Enter employee name -1 to end.\n");
        scanf_s("%s", &emply->emplyName[i]);
        if (strcmp(emply->emplyName[i], "-1") == 0){
            break;
        }

        printf("Enter employee hours.\n");
        scanf_s("%f", &emply->emplyHours[i]);
        printf("Enter Hourly rate.\n");
        scanf_s("%f", &emply->emplyRate[i]);
    }
}
void calculations(input *emply)/*Write a method that calculates the gross, base and overtime pay, pass by reference.*/
{
    int i;
    i = 0;
    for (i = 0; i < 5; i++){
        if (emply->emplyHours[i] > 40) {
            emply->emplyOvrt[i] = (emply->emplyHours[i] - 40) * (emply->emplyRate[i] * 1.5);
        }
        emply->emplyGross[i] = (((emply->emplyHours[i])*(emply->emplyRate[i])) + emply->emplyOvrt[i]);
        emply->emplyBase[i] = (emply->emplyGross[i]) - (emply->emplyOvrt[i]);
        emply->emplyTax[i] = ((emply->emplyGross[i])*.2);
        emply->emplyNet[i] = (emply->emplyGross[i]) - (emply->emplyTax[i]);
        emply->emplyTotal[0] += emply->emplyGross[i];
    }


}


void print(input *emply)
{
    int i;
    for (i = 0; i < 5; i++)
    {
        if (strcmp(emply->emplyName[i], "-1") == 0){
            break;
        }
        printf("Employee Name:%s\n", emply->emplyName[i]);
        printf("Hours Worked:%.2f\n ", emply->emplyHours[i]);
        printf("Hourly Rate:%.2f\n", emply->emplyRate[i]);
        printf("Gross Pay:%.2f\n", emply->emplyGross[i]);
        printf("Base Pay:%.2f\n", emply->emplyBase[i]);
        printf("Overtime Pay:%.2f\n", emply->emplyOvrt[i]);
        printf("Taxes Paid:%.2f\n", emply->emplyTax[i]);
        printf("Net Pay:%.2f\n", emply->emplyNet[i]);
    }
    printf("Total paid to all employees : %.2f\n", emply->emplyTotal[0]);
}
void editEmployees(input *emply){
    int j;
    int index = 1;
    int i;
    printf("Choose an employee.");
    for (j = 1; j < 5; j++); {
        printf("%d.%s", index, emply->emplyName[j]);
    }
    scanf_s("%d", &i);
    employeeInfo(emply);

}
void menu(void){
    printf("Main Menu\n");
    printf("1. Add Employee\n");
    printf("2. Edit Employee\n");
    printf("3. Print Employee\n");
    printf("4. Print All EMployees\n");
    printf("0. exit\n");

}

Answer 1

首先

1. edit employees function: redefinition; different basic types.

您需要转发声明函数原型(prototype)。

2. menu(&payroll) cannot convert from 'input*' to 'input'.

您的函数声明和定义不匹配。

• 在声明中，您有void menu(input payroll);
• 在定义中，您有 void menu(void){//
• 您使用menu(&payroll);调用

以上三个都是不同的。坚持任何一个。

3. employeeInfo(&payroll) cannot convert from 'input*' to 'input'.

同样，函数声明和定义之间不匹配。更改您的函数声明

void employeeInfo(input payroll);

至

void employeeInfo(input* payroll);