Struct isn't readable in AnotherSource but in Source

Question

I have a structure declared in header and implemented in source. Source file can perform on my struct variable, but my main can't see it. Why ?

A.hpp

#include "stdafx.h"
#include <iostream>
#include <stdlib.h>

class A {
public: 
    struct myStruct {
        int someInt;
    } myStructVar;

    void loadA();
};

A.cpp

#include "stdafx.h"
#include "A.h"

void A::loadA() {
    myStructVar.someInt = 10;
    std::cout << myStructVar.someInt << std::endl; // 10 OK
}

Main.cpp

#include "stdafx.h"
#include "A.h"

int _tmain(int argc, _TCHAR* argv[])
{
    A a;
    a.loadA();
    std::cout << myStructVar.someInt() << std::endl; // Error: identifier "myStructVar" is undefined
    system("PAUSE");
    return 0;
}

Error is is Main.cpp

Error: identifier "myStructVar" is undefined.

Thanks.

`myStructVar` is not present in a global scope, but rather in a scope of `A`. Use `a.myStructVar` to reference it. — Algirdas Preidžius, Mar 29 '17 at 14:19
When you fix the first error, you will get more errors. What is `someInt`? — Some programmer dude, Mar 29 '17 at 14:20
Regarding the first problem (that you ask about), perhaps you should [find a good beginners book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) and read before continuing? — Some programmer dude, Mar 29 '17 at 14:21
@Someprogrammerdude Well, it's a member variable of said, inner, `struct`. And, since `A::loadA()` is in `A` scope - it can reference it :) — Algirdas Preidžius, Mar 29 '17 at 14:21
@AlgirdasPreidžius I mean, it's not something that can be *called* :) — Some programmer dude, Mar 29 '17 at 14:21
@DavidPham Yes, it's a member *variable*. Not a member function. — Some programmer dude, Mar 29 '17 at 14:22
@Someprogrammerdude Oh, you're talking about `myStructVar.someInt()` in `main`. Sorry, missed that. :/ — Algirdas Preidžius, Mar 29 '17 at 14:22

score 2 · Accepted Answer · answered Mar 29 '17 at 14:22

2

Change

std::cout << myStructVar.someInt() << std::endl;

To

std::cout << a.myStructVar.someInt << std::endl;

answered Mar 29 '17 at 14:22

Struct isn't readable in AnotherSource but in Source

1 Answers1