I solved a thing, don't know the reason behind

Question

I did a program that translates number into roman numbers, and I was about to show it to a friend of mine when the .exe file didn't work. It asked the user which number to put in, and after he put it the program closed. What I don't understand is why it did work after compiling on DevC++, but had a different behavior as a .exe

I found few solutions out there, and the one that worked for me was to add a:

int a;
cin>>a;

Before returning a cero. Now it works. I don't understand how the console doesn't execute the actions given without that. I'm leaving the code here.


#include <iostream>
using namespace std;

int main(){
    int numero, unidades, decenas, centenas, millares;
    int comprobante;
    
    cout<<"Este programa traduce un numero de 4 cifras a numeros romanos.\nDigite un numero de 
    maximo cuatro cifras: ";
    cin>>numero;
    
     
    unidades= numero%10; 
    numero /= 10; 
    decenas= numero%10; 
    numero/=10; 
    centenas= numero%10; 
    numero/=10; 
    millares=numero;
    comprobante=millares/10; 
        
    if(comprobante==0){
        
        cout<<"Su numero traducido a numeros romanos es: ";
        
        switch(millares){
            case 1: cout<<"M"; break;
            case 2: cout<<"MM"; break;
            case 3: cout<<"MMM"; break;

        }
        
        switch(centenas){
            case 1: cout<<"C"; break;
            case 2: cout<<"CC"; break;
            case 3: cout<<"CCC"; break;
            case 4: cout<<"CD"; break;
            case 5: cout<<"D"; break;
            case 6: cout<<"DC"; break;
            case 7: cout<<"DCC"; break;
            case 8: cout<<"DCCC"; break;
            case 9: cout<<"CM"; break;      
        }
        
        switch(decenas){
            case 1: cout<<"X"; break;
            case 2: cout<<"XX"; break;
            case 3: cout<<"XXX"; break;
            case 4: cout<<"XL"; break;
            case 5: cout<<"L"; break;
            case 6: cout<<"LX"; break;
            case 7: cout<<"LXX"; break;
            case 8: cout<<"LXXX"; break;
            case 9: cout<<"XC"; break;      
        }
        
        switch(unidades){
            case 1: cout<<"I"; break;
            case 2: cout<<"II"; break;
            case 3: cout<<"III"; break;
            case 4: cout<<"IV"; break;
            case 5: cout<<"V"; break;
            case 6: cout<<"VI"; break;
            case 7: cout<<"VII"; break;
            case 8: cout<<"VIII"; break;
            case 9: cout<<"IX"; break;  
        }
    }
    else{
        cout<<"Te dije de 4 cifras. Por desgraciado me cierro.";
    }
    
    int a;
    cin>>a;
    return 0;
}

[Relevant Q with no accepted A](https://stackoverflow.com/q/50346184/7665043), can't be marked as a duplicate because of that. You had no return statement but declared int main which expects something to be returned, so the compiler had to just wing it and it was left broken but built the exe anyways. — l3l_aze, Sep 09 '20 at 17:10
I don´t understand, I indeed had a return 0; statement before. What I added was exclusively int a,; cin>>a; — pemarting, Sep 09 '20 at 17:21
No. Sorry. Probably behavior dependent on the compiler you used and/or the settings, but IDK enough about C++ to be able to help with those. — l3l_aze, Sep 09 '20 at 17:42

score 2 · Answer 1 · answered Sep 09 '20 at 18:14

2

The program closes immediately in the terminal when the program reaches completion. When you cin>>a it is waiting for input before closing. See this.

answered Sep 09 '20 at 18:14

Shane Reilly

251
1
5

Ok, but why it doesn't end when it reaches the return 0? – pemarting Sep 09 '20 at 18:27
After you feed `cin` with input and hit enter, does it exit? – Shane Reilly Sep 09 '20 at 18:29
Yes it does, as expected. – pemarting Sep 09 '20 at 18:44
1

Right, so once it reaches `return 0` it quits the program. Adding in `cin>>a` delays this, which is why you can see it. When you don't have that, the program exits too fast for you to see it. – Shane Reilly Sep 09 '20 at 18:48
Oh so it does the job but closes before you can see it? Alright I get it. Isn't there any other more 'official' solution than typing a cin? Is it the getch they told me in the other answer? – pemarting Sep 09 '20 at 19:25
`getch()` would work, or you can execute your program from within a terminal you've opened yourself. That way, it won't close upon program completion. – Shane Reilly Sep 09 '20 at 19:29
Oh ok, perfect! Appreciated! – pemarting Sep 09 '20 at 19:36

score 0 · Answer 2 · answered Sep 09 '20 at 18:24

0

Include #include<conio.h>

And add getch(); at the end of your code. This will make your console to not close.

answered Sep 09 '20 at 18:24

Marius Pantazi

1
1

Aha, but why is that necessary? – pemarting Sep 09 '20 at 18:27

score 0 · Answer 3 · answered Sep 09 '20 at 21:10

Your code was totally fine.

It is that when the main() reached return 0; that means it is done. And if you are running it as an .exe file, that means the program is done. And when a program is done, it would be closed.

There are couple ways to make it stay not closed. One that many have mentioned is to add a cin >> something; or getch(); with #include <conio.h> for windows at the end, so it would wait till you enter something.

Also you could delay the program before it reaches the return statement. You can check here How do you add a timed delay to a C++ program?

You could also wrap you entire program with a while loop, like:

while(true)
{
    // Your programs here
    if(someCondition) break;
};

where someCondition is a bool type, maybe char c; cin >> c; bool someCondition = (c == 'x'); So that the program only ends if someone input 'x', else it would run again.

I solved a thing, don't know the reason behind

3 Answers3