Infinite loop when typing a character

Question

I’m doing error tests on my "Monkey Test" program and it goes into loop infinite when I type a character on the keyboard other than number.

int ActionGenerator() {

    bool Verify;
    int Action;
    Verify = true;

    while(Verify){
        cout << "\nSELECIONE UMA ACTION: " << endl;
        cout << "[1 = ActionX]\n[2 = ActionZ]" << endl;
        cin >> Action;

        if (Action == 1 || Action == 2){
            return Action;
            Verify = false;

        }else {
            cout << "\nACTION INVALIDA !" << endl;
        }
    }
}

Answer 1

I did another test and the original code I posted doesn’t work in any situation. I did another one which is more complicated but works. It picks up a string and tries to convert it to int. If it fails it is the same as if the number is invalid. Before it works when there was a specific error, it now handles any invalid entry. And as it is done through specialized function, do not need to worry about the state of stream pattern.

#include <iostream>
#include <string>
#include <sstream>

using namespace std;

int ActionGenerator() {
    int Action;
    string Input = "";
    while(true) {
        cout << endl << "SELECIONE UMA ACTION: " << endl;
        cout << "[1 = ActionX]" << endl << "[2 = ActionZ]" << endl;
        getline(cin, Input);
        stringstream myStream(Input);
         if ((myStream >> Action) && (Action == 1 || Action == 2)){
            return Action;
        } else {
            cout << endl << "ACTION INVALIDA !" << endl;
        }
    }
}

int main() {
    cout << ActionGenerator() << endl << "ok" <<endl;
    return 0;
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Note that I did a test with a string more than one character. This failed in the previous code.

---

The code is hard to understand because it is not indented wrong. It may sound silly but it makes a huge difference to understand the code.

There is part that will never be executed. And in fact there is no logic to get out of loop unless you type 1 or 2. And in these cases the execution of the function.

But the same problem is that you are not releasing the state of stream. When an error occurs a flag error is arrow to let you know that there was a problem when you tried to pick up a valid amount. When you type something that cannot be converted into the expected type, in case a int, in addition to the number being converted to 0, is flag stays arrow and only if you explicitly clean up can you go back to work with the stream. If the flag is arrow the stream doesn’t work.

Another problem occurs that the typed data is still in the buffer and you need to ignore what’s in it.

That is, the stream state guard and you have to settle this.

I took advantage and made an improvement to this case:

#include <iostream>
using namespace std;

int ActionGenerator() {
    int Action;
    while(true) {
        cout << endl << "SELECIONE UMA ACTION: " << endl;
        cout << "[1 = ActionX]" << endl << "[2 = ActionZ]" << endl;
        cin >> Action;
        if (Action == 1 || Action == 2){
            return Action;
        } else {
            cout << endl << "ACTION INVALIDA !" << endl;
            cin.clear(); //Limpa a flag de erro quando há falha no parse do valor entrado
            cin.ignore(); //Limpa o buffer
        }
    }
}

int main() {
    cout << ActionGenerator() << endl << "ok" <<endl;
    return 0;
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

There are other ways to treat this but I think for the simple example you gave there’s no reason to do anything more complex than this.

Note that I simplified the loop also. There was no reason for that control in this case.

Answer 2

int ActionGenerator() {

    int opcao;
    char *CleanBuffer;


    do{
        cout << "\nSELECIONE UMA ACTION: " << endl;
        cout << "[1 = ActionX]\n[2 = ActionZ]" << endl;
        scanf("%d", &opcao);
        scanf("%c", &CleanBuffer);

        if (opcao = 1) {
            return opcao;


        }else if(opcao = 2){

            return opcao;

        }

        else {
            cout << "\nACTION INVALIDA !" << endl;
        }

    }while(opcao != 3)


    return 0;
}

Answer 3

The command cin>> does not treat invalid entries. Action is an integer, if someone types a string, generates an untreated exception, which ends up generating the infinite loop. You have to generate code to handle this exception. You have to capture an array type input (char[]), convert it to numbers, with exception treatment for invalid entries, type:

if ( char[0] != '1' )
    ThrowException(); //Seu método para tratar e exceção
else
    number += 1 * (decimalPlace ); // primeira casa decimal neste caso.

Observing: verify = false after return Action will never be executed.