Question

我需要两部分的帮助。该程序的要点是它创建一个菜单并采用L，P或Q的字符输入。＆＃34; L＆＃34;启动一个函数，确定给定年份是否为闰年。＆＃34; P＆＃34;启动一个函数，确定给定的输入是否为素数。＆＃34; Q＆＃34;退出程序。该计划的目的是测试我们调用函数的能力。

第一个问题是我无法确定将返回语句放在何处。将它们放在每个函数的末尾以便返回主菜单似乎是合适的，但这不起作用。这些函数只是保持循环。
第二个问题是，只要在leapYear函数或checkPrime函数中输入字符，就会出现无限循环。例如：进入＆＃39;，＆＃39;或者＆＃39; o＆＃39; o到提示创建一个无限循环的错误消息。是否有办法使我能够防止非数字输入以避免这种情况？
我为格式化道歉。我烦躁不安，试着让它也起作用。猜猜这不是我的一天。

// Project 12B

include <iostream>
include <string>
using namespace std;

 void displayMenu();
 void checkLeapYear();
 void checkPrime();
 char option = 'd';

// Beginning of the main function
int main(int argc, const char * argv[]) {

    // User input dialog
    cout << "Hello! This program provides two different services. "
        << "Entering the character 'L' will initiate a program that can determine if a"
        << " given year is a leap year.\n"
        << " Entering the character 'P' will initiate a program that will determine if a"
        << " given number is prime or not.\n "
        << "Lastly, entering 'Q' will quit the main program.\n\n "
        << "Please enter one of the following characters: L, P, or Q.\n\n";

    cin >> option;


    while (option != 'Q' || option != 'q'){
        displayMenu();
    }

    return 0;

   }


// End of main function




// The displayMenu function

void displayMenu(){

    // Switch statement for menu purposes
    switch (option) {

    case 'p':
    case 'P':
        checkPrime();
        break;

    case 'l':
    case 'L':
        checkLeapYear();
        break;

    case 'q':
    case 'Q':
        cout << "Have a nice day!";
        break;

    default:
        cout << "ERROR: You have entered a value beyond the specified range. Please limit your entries to either"
            << " L,P, or Q. \n\n";
        cin >> option;
        break;
    }
    return;
    }

// The checkLeapYear function

void checkLeapYear(){

    // Declaration of variables for checkLeapYear
    int lOptionYear;

    // User Dialog

    cout << "Hello! This part of the program determines if the year you entered is a leap year."
        << "Please enter a year.\n\n";

    cin >> lOptionYear;

    // Verification that the year entered is actually a valid year

    if (lOptionYear <= 0) {
        cout << "ERROR: You have entered an invalid year.\n\n";
    }

    // Actual leap year verification.
    if ((lOptionYear % 4 == 0 && !(lOptionYear % 100 == 0)) || lOptionYear % 400 == 0) {
        cout << "The year you have entered is a leap year. \n";
    }

    else {
        cout << "The year you have entered is not a leap year\n";
    }
    return;
   }

// The checkPrime function

void checkPrime(){

    // Declaration of variables for checkPrime
    int uNumber,
        factorCount = 0;

    // User dialog
    cout << "Please enter a natural number. The output will be the factors (if it is nonprime) of your input, and the "
        << "total count of factors (if they exist).\n\n";
    cin >> uNumber;

    // Guard against bad input
    if (uNumber <= 0) {
        cout << "You have entered a non-positive number. Please enter a positive, natural number \n\n";
    }

    // The loop starts at two and continues for as long as the iteration-variable is less than or equal to the input number
    for (int k = 2; k < uNumber; k++) {
        if (uNumber % k == 0) {
            cout << k << " ";
            factorCount++;
        }

    }
    // Statement telling total number of factors found
    if (factorCount > 0){
        cout << "\nThere were " << factorCount << " factors found. \n";
    }
    if (factorCount == 0){
        cout << "\nNOTE: The number you entered is prime. \n";
    }
    return;
    }

Answer 1

问题1 ：无论如何都会返回void函数，因此除非您想在中途终止函数，否则无需放置return或担心它们。

问题2 ：

当cin遇到输入时，它无法正确读入指定的变量（例如将字符输入到整数变量中），它会进入错误状态并将输入留在缓冲区中。

您必须做好几件事才能正确处理这种情况。

您必须测试此错误状态。
您必须清除错误状态。
您必须或者处理生成错误状态的输入数据，或者将其刷新并重新提示用户。

因此，在功能checkleapyear中，将cin >> lOptionYear;替换为

while(!(cin >> lOptionYear)){   //test for error state
        cin.clear();            //clear the error state
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
        cout << "Invalid input.  Try again: ";    //repromt user
    }

注意：别忘了添加<limits>

问题3

在main（）中添加 while(option != 'Q' || option != 'q')是不够的，因为输入可能不是l，p或q。取代在main中输入cin>>option，将其带入displayMenu()，并在默认情况下递归调用它：

void displayMenu(){

    cin >> option;   // added here
    // Switch statement for menu purposes
    switch (option) {

    case 'p':
    case 'P':
        checkPrime();
        break;

    case 'l':
    case 'L':
        checkLeapYear();
        break;

    case 'q':
    case 'Q':
        cout << "Have a nice day!";
        break;

    default:
        cout << "ERROR: You have entered a value beyond the specified range. Please limit your entries to either"
            << " L,P, or Q. \n\n";
        //cin >> option;
        displayMenu();   //recursive call
        //break;
    }
    return;
    }

注意：从main移除while（），然后只需调用displayMenu()

错误地放置return语句和character-int不匹配校对

1 个答案: