Question

我试图对＃34;游戏＆＃34;进行介绍，并在其功能中我做了一些是/否，1/2/3，情况。我是新手，但这并不困难，完美无缺。处理无效输入时出现问题。所以这就是现在代码的样子：

#include "Introduction.h"
#include "GameConstants.h"
#include "PlayerCharacter.h"
#include <iostream>
#include <windows.h>

using namespace std;

Introduction::Introduction()
{

}

/////////Function N.1///////////
void Introduction::presentation()
{
    char confirm;
    string enteredName;

    cout << constants.line() << "Welcome traveler! What is the name?" << endl;
    getline(cin,enteredName);// Gets the WHOLE LINE as the name.

    while (confirm != 'Y') //If the player doesn't confirm the name with 'Y' in will run again until it does.
    {
        cout << constants.xline() << "Your name is " << enteredName << " right? (Y/N)" << endl;
        cin >> confirm; //The player's answer
        cin.sync(); //Only takes the first character
        confirm = toupper(confirm); //Turns player message into CAPS for easier detection in the "if" statements

        if (confirm == 'N'){ //If not the correct name, gives another chance
            cout << constants.xline() << "Please, tell me your name again..." << endl;
            cin >> enteredName;
            cin.sync();}

        if ((confirm != 'Y')&&(confirm != 'N')){ //If an invalid input is entered, gives another chance. And insults you.
            cout << constants.xline() << "Fool Go ahead, just enter your name again." << endl;
            cin >> enteredName;
            cin.sync();}
        }

    if (confirm == 'Y'){ //When the answer is yes ('Y') /* Uneeded line */
        PC.setName(enteredName); //Saves the name
        cout << constants.xline() << "Excellent! I have a few more questions for you " << PC.name() << "..." << endl;
    }
}


//////////Function N.2///////////
void Introduction::difSelection(){
    int selectedDif = 0; //Variable to store selected difficulty whitin this function.

    Sleep(2500);

    cout << constants.xline() << "What kind of adventure do you want to take part in?" << endl;

    Sleep(2500); //Wait 2,5 s

    cout << "\n1= Easy\n2= Normal\n3= Hard" << endl;

    while(selectedDif != 1&&2&&3){ //Selected option must be 1/2/3 or will run again
        cin >> selectedDif; //Sets the user selected difficulty
        cin.sync(); //Gets only first character 
        if((selectedDif != 1||2||3)&&(!(selectedDif))){ //If the input isn't 1/2/3 AND is an invalid character, this will run. And it'll start again

            cout << constants.xline() << "Criminal scum. Go again." << endl;
            cin.clear();
            cin.ignore();
        }

        if(selectedDif != 1&&2&&3){ //If selected option isn't 1/2/3, this will run and will loop again. However I know this conflicts with the previous statement since this will run anyways.

        cout << constants.xline() << "Wrong input, please try again." << endl;
        } 
        else if(selectedDif == 1){
        constants.setDiff(1);
        constants.setStatPoints(15);
        } else if(selectedDif == 2){
        constants.setDiff(2);
        constants.setStatPoints(10);
        } else if (selectedDif == 3){
        constants.setDiff(3);
        constants.setStatPoints(5);}
    }

}

第一个功能完美无缺，您可以键入＆＃34; aaa＆＃34;或者＆＃34; a a a＆＃34;并将工作。但是，我想知道是否有更简单的方法。（初学者可以理解，3天前刚开始大声笑;如果它包含一些高级或不太知名的代码，现在希望保持这样）。

现在，第二个，我真的不知道如何解决它。我需要一些东西，如果用户的输入是无效的字符类型，抛出某个消息，如果它是一个int类型，但超出范围，另一个消息。当然，如果失败则再次运行。经过大量搜索，无法找到符合此要求的任何内容。

Answer 1

要检查用户输入是否为int，您可以使用good()函数。

int val;
cin >> val;

if( cin.good() ) {
    // user input was a valid int
} else {
    // otherwise
}

对于范围检查，语法略有不同。如果数字不等于1也不是2也不是3：

，则返回true

selectedDif != 1 && selectedDif != 2 && selectedDif != 3

另一种较短的方法是使用：

selectedDif < 1 || selectedDif > 3

另一件事，在c ++中，有两个关键字break和continue，它们可以减少循环中的代码。

安全[Y / N]; [1/2/3 /等]功能

1 个答案: