Working Outside of Functions

[CaleyM]

Is it invalid to define variables and use i/o outside of a function? When I tried compiling this it complains about all lines except the bold declarations: "does not define a type". I have no clue what that means.

corporation Corporation1;
Corporation1.wealth = 1000; Corporation1.sharePrice = 1;
person Player1;
Player1.wealth = 1000;
cout << "Choose corporation name: ";
cin >> Corporation1.name;
cout << "Corporation named: " << Corporation1.name;

[Texan40]

It means that it doesn't recognize "corporation" or "player" as data types. You may need to include the header defining them?

[therockon7throw]

Each variable has a scope, and they are valid only in their scope.
A local variable is only valid in the scope it was defined, a global variable is valid in all part of the program.

[CaleyM]

The corporation and person structs were defined directly above that at the root level. The 5 lines moved back into main() compiles:

// Begin: Structs
struct corporation {...};
struct person {...};
struct share{...};
// End: Structs
// Begin: Objects
corporation Corporation1;
person Player1;
// End: Objects
void MenuBuySell1 (corporation Corporation) {...}
int main()
{
    ...
    cout << "Choose corporation name: ";
    cin >> Corporation1.name;
    cout << "Corporation named: " << Corporation1.name;
    Corporation1.wealth = 1000; Corporation1.sharePrice = 1;
    Player1.wealth = 1000;
    ...
}

So, I guess there's a rule that operators have to be in functions?

[guestgulkan]

You are not allowed to have free standing code in the space between functions.
That space is for global variable definitions, and declarations and function definitions and such like.

[vlad from moscow]

You should show the exact text of error messages.