如何将信息从循环存储到数组中

问题描述 投票:-3回答:2

我需要输入信息并为几个员工计算一些东西,并将每个员工信息输出到单个控制台,我需要使用数组。

问题是我真的不知道如何将循环中的信息存储到数组中。运动的屏幕截图是here

我问用户有多少工人,并且值进入“workers”变量,然后我创建int employees [workers]数组,因此循环中的迭代次数由用户的输入决定。

我的循环问题在于,无论有多少员工,它都不会重复问题。

我使用do while循环和“Count”变量来控制重复次数,但在输入信息一次后,它只显示结果,而不是再次询问问题。

我也尝试了while循环和“count”变量,但这次它只询问有多少员工,它只显示空输出。

int main()
{
//************************DECLARATIONS**********************

    typedef char INFO;

    INFO f_name[30];
    INFO m_name[10];
    INFO l_name[30];

    int count; // tracks the number of iterations in do loop
    int workers;
    double rate;
    double hrs_worked;
    double gross_inc;
    double overtime;
    double tax_total;
    float net;

    float STATE_TAX;
    float FED_TAX;
    float UNI_FEE;
    const double OVERTIME_R = 1.5;

//*****************INPUT FROM USER***************************
    cout << "Enter amount of workers" << endl;
    cin >> workers;

    int employees[workers];

    while(count < workers)
    {

        cout << "Enter worker's First name: " << endl;
        cin.getline(f_name, (sizeof(f_name)-1));

        cout << "Enter worker's middle name initial: " << endl;
        cin.getline(m_name, (sizeof(m_name)-1));

        cout << "Enter worker's last name: " << endl;
        cin.getline(l_name, (sizeof(l_name)-1));

        cout << "Enter number of hours worked: " << endl;
        cin >> hrs_worked;

        // If statement activates if user enters incorrect values
        // and asks to reenter the correct value.
        if(hrs_worked < 0 || hrs_worked > 60)
        {
            while(hrs_worked < 0 || hrs_worked > 60)
            {
                cout << "Must be between 0 and 60: " << endl;
                cin >> hrs_worked;
            }
        }

        cout << "Enter Rate Per Hour: " << endl;
        cin >> rate;

        // If statement activates if user enters incorrect values
        // and asks to reenter the correct value.
        if(rate < 0 || rate > 50)
        {
            while(rate < 0 || rate > 50)
            {
                cout << "Must be between 0 and 50: " << endl;
                cin >> rate;
            }
        }
        count++;
    }
    system("clear");
//**********************************CALCULATIONS*****************
    // Calculates overtime if employee worked more than 40 hrs

    if(hrs_worked > 40)
    {
        overtime = (hrs_worked - 40.0) * rate * OVERTIME_R;
    }

    gross_inc = (rate * hrs_worked) + overtime;
    STATE_TAX = gross_inc * 0.06;
    FED_TAX = gross_inc * 0.12;
    UNI_FEE = gross_inc * 0.02;
    tax_total = STATE_TAX + FED_TAX + UNI_FEE;
    net = gross_inc - (tax_total)


    return 0;
}   

目前,优先级是设置正确的循环并将信息从循环存储到数组中。此时输出不是主要关注点。

c++ arrays loops
2个回答
1
投票

第一个问题是你需要了解这一行中发生的事情:

    int count; // tracks the number of iterations in do loop

您只需输出此变量的值:

    cout << "count: " << count << endl;

之后你应该阅读一些关于在C ++中定义和声明变量的内容(它与“操作”不同)。

如果您只需要一个简单的解决方案,我可以告诉您,您的count变量将具有一些“垃圾”值。解决它的最简单方法是将其初始化为起始值。在分析了上下文后,我可以假设int count = 0;将是正确的起始值。

我想可能还有其他问题,但这个问题与你的问题直接相关。

祝你好运C ++!我还建议你深入研究C ++基础知识,首先要了解变量的定义和声明。

[UPDATE]

我想在您发布更新后添加一些内容:

    int i; // will be used in for loop

请不要这样做。如果你想在“for循环”中使用它,那么在那里初始化它。

    int workers = 0;

如果此变量表示“工人数量”,则应根据其含义对其进行命名,例如:

    int numberOfWorkers = 0;

你的问题来自这个li(n)e(s):

    // typedef for string data types    
    typedef char INFO;

不幸的是,你在这个评论中撒谎。这不是string类型。你的typedefchar的别名所以它只存储一个字符,而不是string

它应该是

    // typedef for string data types    
    typedef char* INFO;

但是在我的拙见中,由于INFO的类型名称没有说什么,所以它是不情愿的。此外,你必须记住,如果你想解决这个问题,你还必须为这些c-string成员(f_namem_namel_name)设置一些固定的大小,因为c-strings意味着具有恒定的大小。

还有另外一种解决方案 - 如果你想用C ++编写代码,更喜欢std::string而不是c-strings。简而言之,std::string的工作方式类似于char类型元素的动态大小数组/容器。

你也可以简单地替换c风格的“动态”数组:

    struct employee *emp = new employee[workers];

与...的std ::向量:

    std::vector<employee> emp;
    emp.reserve(workers);

(此行中也不需要使用struct关键字)。

所有输入检查都会发生另一个错误,例如:

    while(emp[i].hrs_worked < 0 || emp[i].hrs_worked > 60)
    {
        cout << "Must be between 0 and 60: " << endl;
        cin >> emp[i].hrs_worked;
    }

它导致无限循环(我检查了它)。为什么?因为你没有清理输入缓冲区,例如:How do I flush the cin buffer?

我认为您也可以考虑将此结构更改为类,并考虑为您的类重载i / o流运算符(operator <<和operator >>),这将简化employee对象的i / o操作。

最后一句话 - 请通过编辑上一个问题而不是将其作为答案来发布您的更新。

再一次 - 祝你学习C ++好运!


0
投票

[更新]基本上,代码是垃圾,但它让我记住并学习了很多东西。所以这就是我重新编写代码的方式:

我使用了struct,pointer和“new”运算符。

唯一有问题的部分是这个输入部分,因为如果我输入多个字符,程序基本上会跳过所有内容,只显示一个模板:

cout << "Enter first name of employee    "<<i+1<<" : " << endl;
cin  >> emp[i].f_name;

cout << "Enter middle name of employee   "<<i+1<<" : " << endl;
cin  >> emp[i].m_name;

cout << "Enter last name of employee     "<<i+1<<" : " << endl;
cin  >> emp[i].l_name;

cout << "Hours of work by employee " << i+1 << ": " << endl;
cin >> emp[i].hrs_worked;

这是输出的样子:output

//*********************** Preprocessor Directives *********************
#include <fstream>
#include <iostream>
#include <iomanip>
#include <string.h>
#include <cmath>
#include <stdlib.h>
#include <string.h>

using namespace std;

// Constants
const double OVERTIME_R = 1.5;
#define STATE_TAX 0.06
#define FED_TAX 0.12
#define UNION_FEE 0.02

// typedef for string data types
typedef char INFO;

// using struct to store employee info
struct employee
{
    INFO f_name;
    INFO m_name;
    INFO l_name;
    float rate;
    float hrs_worked;
    float gross;
    float overtime;
    float state_tax;
    float fed_tax;
    float uni_fee;
    float net;
};

//******************************* main ********************************

int main()
{

    int workers = 0;
    int i; // will be used in for loop
    float total_gross = 0;
    float avg_gross = 0;

//*****************Process***************************

    // asking number of employees
    cout << "Enter the number of employees: " << endl;
    cin >> workers;

    // array of employees
    struct employee *emp = new employee[workers];

    for(i = 0; i < workers; i++)
    {

        cout << "Enter first name of employee    "<<i+1<<" : " << endl;
        cin  >> emp[i].f_name;

        cout << "Enter middle name of employee   "<<i+1<<" : " << endl;
        cin  >> emp[i].m_name;

        cout << "Enter last name of employee     "<<i+1<<" : " << endl;
        cin  >> emp[i].l_name;

        cout << "Hours of work by employee " << i+1 << ": " << endl;
        cin >> emp[i].hrs_worked;

        // If statement activates if user enters incorrect values
        // and asks to reenter the correct value.
        if(emp[i].hrs_worked < 0 || emp[i].hrs_worked > 60)
        {
            while(emp[i].hrs_worked < 0 || emp[i].hrs_worked > 60)
            {
                cout << "Must be between 0 and 60: " << endl;
                cin >> emp[i].hrs_worked;
            }
        }

        cout << "Rate Per Hour of employee " << i+1 << ": " << endl;
        cin >> emp[i].rate;

        // If statement activates if user enters incorrect   >> values
        // and asks to reenter the correct value.
        if(emp[i].rate < 0 || emp[i].rate > 50)
        {
            while(emp[i].rate < 0 || emp[i].rate > 50)
            {
                cout << "Must be between 0 and 50: " << endl;
                cin >> emp[i].rate;
            }
        }

        // if employee has worked over 40 hrs. this statement activates.
        if(emp[i].hrs_worked > 40)
        {
            emp[i].overtime = (emp[i].hrs_worked - 40.0) * emp[i].rate *         
OVERTIME_R;
        }

        // Calculating the taxes.
        emp[i].state_tax = emp[i].gross * STATE_TAX;
        emp[i].fed_tax = emp[i].gross * FED_TAX;
        emp[i].uni_fee = emp[i].gross * UNION_FEE;
        emp[i].net= emp[i].gross - (emp[i].state_tax + emp[i].fed_tax +                 
  emp[i].uni_fee);

        // Total Gross
        total_gross += emp[i].gross;
    }

//**********************************OUTPUT****************************
    cout << endl;
    cout << endl;
    cout << "\t\t\t\t" <<"Data Housing Corp. Weekly Payroll" << endl;
    cout << "\t\t\t\t" <<"~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~" << endl;
    cout << "First Name" << setw(5) << "MI" << setw(13) << "Last Name" <<     setw(18)
         << "Rate per hour" << setw(18)<< "Hours worked" << setw(12)<< "Overtime"
         << setw(10)<< "Gross" << setw(15)<< "State tax"
         << setw(10)<< "Fed tax" << setw(15)<< "Union fee" << setw(10)<< "Net" << endl;
    cout << "==========" << setw(5) << "==" << setw(13) << "=========" <<     setw(18)
         << "=============" << setw(18)<< "============" << setw(12)<< "========"
         << setw(10)<< "=====" << setw(15)<< "========="
         << setw(10)<< "=======" << setw(15)<< "=========" << setw(10)<< "==="     << endl;

for ( i = 0  ; i < workers ;  i++)
    {
        cout << setw(10) << emp[i].f_name << setw(5) << emp[i].m_name << setw(13)
        << emp[i].l_name << setw(18) << emp[i].rate << setw(18)<<     emp[i].hrs_worked
        << setw(12)<< emp[i].overtime << setw(10)<< emp[i].gross << setw(15)
        << emp[i].state_tax << setw(10)<< emp[i].fed_tax << setw(15)<< emp[i].uni_fee
        << setw(10) << fixed << showpoint <<setprecision(2)<< emp[i].net << endl;
    }
    return 0;
}
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