C++ class template inheritance puzzle

Question

What's wrong with this code? gcc 4.6.1 is complaining "‘foo’ was not declared in this scope" in baz(). If I transform the code so that one of the templates is just a regular class, the problem goes away.

struct Foo {
    char foo;
};

template<int N>
struct Bar : public Foo
{
    Bar() { foo; }
};

template<int N>
struct Baz : public Bar<N>
{
    void baz() { foo; }
};

int main() {
    Baz<10> f;
    return 0;
}

I did not understand what you are trying to do here - Bar() {foo;} What is foo? — badmaash
– badmaash, Commented Oct 31, 2011 at 18:57
This is called dependent name. See womble.decadent.org.uk/c++/template-faq.html#base-lookup — Suma
– Suma, Commented Oct 31, 2011 at 19:00
possible duplicate of Why doesn't this C++ template code compile? — Suma
– Suma, Commented Oct 31, 2011 at 19:02
badmaash, you sound like a compiler. :) foo is the foo from the class Foo, inherited by Bar, of course. — Joshua Chia
– Joshua Chia, Commented Feb 24, 2012 at 22:27

lvella · Accepted Answer · 2011-10-31 18:56:30Z

1

What is wrong, according to the specifications, I don't know, but you may make your code to compile by using:

void baz() { Bar<N>::foo; }

answered Oct 31, 2011 at 18:56

lvella

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1 Comment

Mooing Duck Over a year ago

What is wrong, is that while compiling Baz it doesn't know which Bar<N> to inherit from yet (there might be specializations later), so it doesn't yet know there is a foo member. By qualifiying it, you're telling it that Bar<N> will have the foo member, and not to worry about it until N is known.

Mike Seymour · Accepted Answer · 2011-10-31 19:25:51Z

1

foo is a dependent name; that is, it depends on the template parameter, so until the template is instantiated the compiler doesn't know what it is. You have to make it clear that it is a class member, either Bar<N>::foo or this->foo.

(You probably also want to do something with it; simply using it as the ignored value of an expression doesn't do anything at all).

answered Oct 31, 2011 at 19:25

Mike Seymour

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