我如何继续该功能而不退出c ++中的else语句

Question

Hello so my question is how do I continue on with new inputs after an else statement? In my program, I wrote an else statement that if the input is neither 1 nor 2, the user has to put a new value to get the result he/she wants. But after my else statement, my program shuts down. I don't know how to fix this. I'm very new to c++ so please keep harsh comments to yourself... Here's my code:

// This program will allow the user to input from the keyboard 
// whether the last word to the following proverb should be party or country:
// "Now is the time for all good men to come to the aid of their _______"
// Inputting a 1 will use the word party. A 2 will use the word country.



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


void writeProverb(int);


int main ()
 {

int wordCode;

cout << "Given the phrase:" << endl;
cout << "Now is the time for all good men to come to the aid of their ___" << endl;
cout << "Input a 1 if you want the sentence to be finished with party" << endl;
cout << "Input a 2 if you want the sentence to be finished with country." << endl;
cout << "Please input your choice now" << endl;
cin  >> wordCode;
cout << endl;
    writeProverb(wordCode);

return 0;
}



 void writeProverb (int number)


{
if (number == 1)
    cout << "Now is the time for all good men to come to the aid of their      party." << endl;

else if (number == 2)   
    cout << "Now is the time for all good men to come to the aid of their country." << endl;

else
{
    cout << "I'm sorry but that is an incorrect choice: Please input a 1 or 2"  << endl;
 } 


 }

So basically, after the else statement, I want my program to wait for the user to enter 1 or 2 instead of just shutting down.

Answer 1

do {
    cout << "Please input your choice now" << endl;
    cin  >> wordCode;
    cout << endl;
    writeProverb(wordCode);
}while (wordCode != 1 && wordCode != 2);

This code exits if user inputs 1 or 2. Stays otherwise.

Answer 2

You want to have a do-while construct until you get a legal value, as Sakthi Kumar already pointed out.

However , you do not want to move the knowledge of what a legal value is up to main. Therefore, have the writeProverb method return if the value is legal or not. This keeps things in the proper abstraction level. You should also consider updating the "menu" printing by using an object, thus tying everything together.

// ... in main()
    do {
        cout << "Please input your choice now" << endl;
        cin  >> wordCode;
        cout << endl;
        writeProverb(wordCode);
    } while( !writeProverbIfLegalNumber(wordCode) );
}

bool writeProverbIfLegalNumber (int number) {
    if (number == 1) {
        cout << "Now is the time for all good men to come to the aid of their      party." << endl;
        return true;
    }
    else if (number == 2) {
        cout << "Now is the time for all good men to come to the aid of their country." << endl;
        return true;
    }
    else
    {
        cout << "I'm sorry but that is an incorrect choice: Please input a 1 or 2"  << endl;
    } 
    return false;
}

Answer 3

This is basic logic, you should read about loops and how to use them. This while loop continues while wordCode is not between 1 and 2 inclusive; in other words, it continues until you get 1 or 2.

int main ()
{

    int wordCode;

    cout << "Given the phrase:" << endl;
    cout << "Now is the time for all good men to come to the aid of their ___" << endl;
    cout << "Input a 1 if you want the sentence to be finished with party" << endl;
    cout << "Input a 2 if you want the sentence to be finished with country." << endl;



    do
    {
       cout << "Please input your choice now 1 or 2: " << endl;
       cin  >> wordCode;

    } while(!(1 <= wordCode && wordCode <=2));

    cout <<endl;
    writeProverb(wordCode);

    return 0;
}

Answer 4

in this case,you can put the judging statement in the "main" function. You can use a "while" statement to control the logic.

boolean flag = true;
while(flag){
    cout<<"input";
    cin>>wordCode;
    if(wordCode==1 || wordCode==2){
        flag=false;
        writeProverb(wordCode);
    }
}