Apologies for the most ambiguous and bizarre title.
Suppose we have 2 classes A and B.
class B has interface hasSmth but class A has not.
How to make this code evaluate without compile errors?
class A {
//..implementation
int id() { return 1; }
};
class B {
//..implementation
int id() { return 2; }
bool hasSmth() { return true; }
};
int main()
{
auto obj = someFunction();//returns A or B
if (obj.id() == 1 || (obj.id() == 2 && obj.hasSmth())) {
...
}
}
If the function returns obj of type B, then we are good.
But if it returns obj of type A, compiler will complain about A not having hasSmth, regardless of that part of if never been evaluated.
Can someone give a workaround please?
Can someone give a workaround please?
Read the declaration of someFunction
to see what it returns. In the case it doesn't return B
, then don't write obj.hasSmth()
. Problem solved.
Now, let's change the question a bit. Let's say that you want to make this work without knowing the return type. Perhaps because rather than main
you may be actually writing a template that works with different types. There are several approaches, but function overloads are a simple one:
bool check([[maybe_unused]] const A&) {
return true;
}
bool check(const B& b) {
return b.hasSmth();
}
template<bool returnsA>
void foo() {
auto obj = someTemplate<returnsA>(); // returns A or B
if (check(obj)) {