Member reference base type 'char' is not a structure or union in C++

Question

Can someone help me to debug this problem in C++? I want it to loop if the user want to enter once again, but the error keeps on showing up.

void typeA(){
    int j;
    char dec;
    do{
        cout << "Enter Month: ";
        cin >> month[j];
        cout << "Enter Date: ";
        cin >> date[j];
        cout << "Enter Time in Hour: ";
        cin >> aHours[j];
        cout << "Enter Time in Minutes: ";
        cin >> aMins[j];
        cout << "Enter Time Out Hour: ";
        cin >> aHours[j];
        cout << "Enter Time Out Minutes: ";
        cin >> aMins[j];

        cout << "Enter Again?: [y/n]";
        cin >> dec;

    }while(!dec.compare('y'));
    cout << "Exit";

}

Hi Raven! Could you please paste absolutely all of your code, even `#include` statements that might not seem important. (If you don't have any `#include` statements, that would be your problem right there.) — Arthur Tacca, Aug 29 '18 at 14:50
`!dec.compare('y')` char has no `operator.` because it is not a union/struct as the error says. I dont know what you were attempting to do with that. most likely dec != 'y' ? — Borgleader, Aug 29 '18 at 14:51
`char` is a basic type. In C++, basic types can't have methods. You should simply compare `dec != 'y'` — Yksisarvinen, Aug 29 '18 at 14:51
@ArthurTacca No #include will fix trying to call a non-existant `compare` method on a `char` — Borgleader, Aug 29 '18 at 14:51
@Borgleader But it would've helped me paste the code and run it through a compiler and see where the error was without having to pick through it myself. At the moment it will probably complain about unknown symbol `cout`. (Pasting the exact error message including the line number would probably have sufficed too.) Plus it is good practice to let new people know about. — Arthur Tacca, Aug 29 '18 at 14:53
I'm trying to compare the decision of the user. I think my statement if incorrect can you help me to correct it? — Raven, Aug 29 '18 at 14:57
Please **[edit]** your question with an [mcve] or [SSCCE (Short, Self Contained, Correct Example)](http://sscce.org) — NathanOliver, Aug 29 '18 at 14:58
@Raven As the other people stated, char does not have compare (or any other method). Are you mistaking it for string? Simply write `dec != 'y'` as already stated. != and == are also perfectly fine on string, by the way. — Aziuth, Aug 29 '18 at 15:00
@Raven If you think your statement is correct, you should start with [a good C++ book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). C++ is not Java or Python and don't try to draw parallels. `char` has no methods. You can't call a method on `char` variable. — Yksisarvinen, Aug 29 '18 at 15:01
You cannot "debug" compiler errors, you just need to fix them — Slava, Aug 29 '18 at 15:04
@Raven `j` is uninitialized and doesn't change between iterations of the loop. `month`, `date`, `aHours`, `aMins`, `aHours` and `aMins` are undefined. Are you using global variables? If so, stop it. — Swordfish, Aug 29 '18 at 15:13

Mohit · Accepted Answer · 2018-08-29T15:35:31.003

1

You should use == operator to check if two fundamental types are equal or not, because the fundamental types in C++ can't have methods.

void typeA(){
    int j;
    char dec;
    do{
        cout << "Enter Month: ";
        cin >> month[j];
        cout << "Enter Date: ";
        cin >> date[j];
        cout << "Enter Time in Hour: ";
        cin >> aHours[j];
        cout << "Enter Time in Minutes: ";
        cin >> aMins[j];
        cout << "Enter Time Out Hour: ";
        cin >> aHours[j];
        cout << "Enter Time Out Minutes: ";
        cin >> aMins[j];

        cout << "Enter Again?: [y/n]";
        cin >> dec;

    }while(dec == 'y');
    cout << "Exit";

}

edited Aug 29 '18 at 15:35

answered Aug 29 '18 at 15:05

Mohit

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There is no such thing as `operator==` for `char`. – Swordfish Aug 29 '18 at 15:10
@Swordfish Why does that work? – Mohit Aug 29 '18 at 15:14
it works tho. i dont know why but yes it works. – Raven Aug 29 '18 at 15:15
@Mohit new your explanation is not misleading anymore. – Swordfish Aug 29 '18 at 15:15
@Swordfish Sorry, But I mean that. – Mohit Aug 29 '18 at 15:16
@Raven it was just about using the term `operator==` which is a different thing than `==` in C++. – Swordfish Aug 29 '18 at 15:16
More correct term for C++ is fundamental types, not primitive. https://en.cppreference.com/w/cpp/language/types – Slava Aug 29 '18 at 15:33

Member reference base type 'char' is not a structure or union in C++

1 Answers1