Goto function seems to loop the first half of the code

Question

I normally don't create codes for myself, but today (or tonight, depending on your time zone) I created a simple calculator code which lets you enter multiple numbers, not just two items. This was a very ambitious idea, and so I got to work and then I tested it, and when I inserted my two numbers and my operator, it went back to the very first input (fyi, its the "please insert a number" prompt) and asked me again to insert a number, instead of checking my operator like I expected it to work.

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

int main() {
    int ans=0, num=0, secnum=0;
    char oper, option;
    bool statcheck = false;

    while(!statcheck) {
        cout<<"Please insert a number."<<endl;
        cin>>num;
        cout<<"Please insert another number."<<endl;
        cin>>secnum;
        cout<<"Select the operator you would like to use."<<endl;
        cin>>oper;
        switch(oper) {
            case '+':
                ans += num, secnum;
                break;
            case '-':
                ans -= num, secnum;
                break;
            case '*':
                ans *= num, secnum;
                break;
            case '/':
                ans /= num, secnum;
                break;
            case '%':
                ans %= num, secnum;
                break;
            default:
                cout<<"Error."<<endl;
                break;
        }
        goto anotherone;
        cout<<"Would you like to add another number to the equation? (Y for yes, any other number for no)"<<endl;
        cin>>option;
        toupper(option);
        if (option == 'Y') {
            goto there;
            secnum=0;
            cout<<"What number would you like to add to the equation?"<<endl;
            cin>>secnum;
            cout<<"Select the operator you want to use."<<endl;
            cin>>oper;
            switch(oper) {
                case '+':
                    ans += secnum;
                    break;
                case '-':
                    ans -= secnum;
                    break;
                case '*':
                    ans *= secnum;
                    break;
                case '/':
                    ans /= secnum;
                    break;
                case '%':
                    ans %= secnum;
                    break;
                default:
                    cout<<"Error."<<endl;
                    there:;
                    break;

            }
            anotherone:;

        } 
        else {
            cout<<ans;
            statcheck = true;
        }

    }
}

This seems to be the issue of the goto functions that I used in the code, as when I remove them, it works just fine, minus the ability to trap the user from inserting a random input in the "if (option == y)" condition.

I was expecting it to execute perfectly: it would ask for two numbers, the operator, ask if they wanted to add another number into the equation or not, and then it would give the answer. But of course, as an amateur at C++ and coding in general, once I got this unexpected and weird error, my mind was twisted (mainly because I'm very bad at debugging things, to be honest).

`goto` isn't a function, and should generally be avoided in favor of looping constructs. — πάντα ῥεῖ, Apr 16 '23 at 10:46
Step through your code line by line with your debugger to get a grasp what's actually going on. — πάντα ῥεῖ, Apr 16 '23 at 10:51
Avoid goto, you may use [break](https://en.cppreference.com/w/cpp/language/break) to break out of a loop. — Pepijn Kramer, Apr 16 '23 at 10:52
https://stackoverflow.com/questions/46586/goto-still-considered-harmful — Pepijn Kramer, Apr 16 '23 at 10:53
`goto anotherone;` could be replaced with `continue;`, since there are no more statements after the `anotherone` label. Since this is done unconditionally though, any code in the loop body after this `goto` is irrelevant and will never be executed. Avoid mixing pasta with code... — fabian, Apr 16 '23 at 10:54

score 0 · Answer 1 · edited Apr 22 '23 at 10:33

The goto statement passes the control to the label notherone in the inner if statement of the while loop. In fact you have:

while(!statcheck) {
    cout<<"Please insert a number."<<endl;
    cin>>num;
    cout<<"Please insert another number."<<endl;
    cin>>secnum;
    cout<<"Select the operator you would like to use."<<endl;
    cin>>oper;
    switch(oper) {
        case '+':
            ans += num, secnum;
            break;
        case '-':
            ans -= num, secnum;
            break;
        case '*':
            ans *= num, secnum;
            break;
        case '/':
            ans /= num, secnum;
            break;
        case '%':
            ans %= num, secnum;
            break;
        default:
            cout<<"Error."<<endl;
            break;
    }
    goto anotherone;
    //...
    if (option == 'Y') {
        //....
        anotherone:;
    } 
    else {
        cout<<ans;
        statcheck = true;
    }

}

So after passing the control to the label anotherone the while loop executes its next iteration. And almost a half of the program never gets the control. That is what you do is what you get.

Take into account that as the control is passed to the if statement then its else part does not get the control. It is skipped.

It is a bad style of programming to use the goto statement. It makes the code unreadable.

Pay attention to that in statements like that:

ans += num, secnum;

there are used expressions with the comma operator. In fact the shown statement is equivalent to:

ans += secnum;

that does not make sense.

Goto function seems to loop the first half of the code

1 Answers1