[英]C++ array structures
我正在阅读本书中关于结构的章节,它让我重新修改了我已经制作的程序,但这次使用了我以前从未使用过的结构; 然而,在完成该计划后,有一个我不理解的问题。 程序的输出只显示一次。 它处于for循环中,但即使它要求我输入三次信息,它也只输出第一个信息。
我可能只是不了解结构中的数组是如何工作的。 我的问题的一个例子如下。 我在以下循环中输出了输出
for(int counter = 0; counter <size; counter++)
大小为3,这意味着我将打印输出三次; 但是我得到的答案就像我要求的一样。
Listofnames[0].F_name
当我真正想要的是
Listofnames[0].F_name Listofnames[1].F_name Listofnames[2].F_name
但是,我不想写三次,我做了测试它实际上有效,但这是唯一的方法吗? 或者我在程序中遗漏了什么?
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
struct Names
{
string F_name; //Creating structure called Names.
string L_name;
char Mi;
};
struct Payrate
{
double rate;
double hoursworked; //Creating structure called Payrate.
double gross;
double net;
};
int main()
{
double stateTax = 0, federalTax = 0, unionFees = 0, timeHalf = 1.5; //Initializing variables.
const int size = 2; //Array size.
Payrate employee[size]; //Structure variables
Names Listofnames[size];
for (int counter = 0; counter < size; counter++) //Initializing for loop.
{
cout << "What's your first name?: " << endl;
cin >> Listofnames[counter].F_name;
cout << "What's your last name?: " << endl; //Displaying names, and hours worked, rate.
cin >> Listofnames[counter].L_name;
cout << "What is your middle initial?: " << endl;
cin >> Listofnames[counter].Mi;
cout << "How many hours did you work? Please enter a number between 1-50: " << endl;
cin >> employee[counter].hoursworked;
cout << "What is your hourly rate? Please enter a number between 1-50: " << endl;
cin >> employee[counter].rate;
if (employee[counter].hoursworked < 0 || employee[counter].hoursworked >50) //Initializing conditional statements.
{
cout << "Sorry you entered a erong entry. Pc shutting off " << endl; //Displays what happens is user inputs a number under 0 or over 50.
}
if (employee[counter].rate < 0 || employee[counter].rate > 50) //Initializing conditional statements.
{
cout << "Sorry you entered a erong entry. Pc shutting off " << endl; //Displays what happens is user inputs a number under 0 or over 50.
}
if (employee[counter].hoursworked <= 40) //Initializing conditional statements.
{
employee[counter].gross = employee[counter].hoursworked * employee[counter].rate; //Calculating gross.
}
else if (employee[counter].hoursworked > 40) //Initializing conditional statements.
{
employee[counter].gross = employee[counter].hoursworked * (employee[counter].rate * timeHalf); //Calculating gross.
}
stateTax = employee[counter].gross * 0.06;
federalTax = employee[counter].gross * 0.12; //Calculates all the tax fees, and net.
unionFees = employee[counter].gross * 0.02;
employee[counter].net = employee[counter].gross - (stateTax + federalTax + unionFees);
}
cout << "FirstN " << "MI " << "LastName " << "\t" << "Rate " << "HoursWorked " << "TimeHalf " << "StateTax " << "FederalTax " << "UnionFees " << "Gross " << " " << "Net " << endl; //Displays header of output.
cout << "==================================================================================================================" << endl;
for (int counter = 0; counter <= size; counter++)
{
//Output.
cout << Listofnames[counter].F_name << "\t" << fixed << setprecision(2) << Listofnames[counter].Mi << " " << Listofnames[counter].L_name << "\t" << employee[counter].rate << "\t" << employee[counter].hoursworked << "\t" << setw(7) << timeHalf << "\t" << setw(8) << stateTax << setw(12) << federalTax << "\t" << unionFees << "\t" << employee[counter].gross << "\t" << employee[counter].net << endl;
system("pause");
}
}
Ps如果你不得不重新修改这个程序,你会用什么来简化它。 要求我可以继续重新修改,并学习更高级的东西。 矢量,指针? 提前致谢。
你有一个包含3个索引的数组,但你的循环只能达到2个索引。 将for循环更改为此。
for (int counter = 0; counter <= size; counter++)
现在,这个循环将打印所有索引。
您也可以使用它来代替使用静态值。
for (int counter = 0; counter < sizeof(Listofnames)/sizeof(Listofnames[0]); counter++)
sizeof(Listofnames)/sizeof(Listofnames[0]) This will give you the total size of your array.
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