How to initialize members of template-base class from derived constructor initializer list?

Question

This:

template <typename T>
struct base {
    T a;
};

template <typename T>
struct derived : base<T> {
    derived(T v) : a(v) {} // xxx: how?
};

int main() {
    return 0;
}

Goes boom:

test.cc: In constructor ‘derived<T>::derived(T)’:
test.cc:12:20: error: class ‘derived<T>’ does not have any field named ‘a’
     derived(T v) : a(v) {} // xxx: how?

If I replace the a(v) with { this->a = v; } it's fine, is there any way to initialize members of a templated base class from the initializer list of a derived class?

You can't, and this has nothing to do with templates. A class can only initialize its own class's members. As far as the base classes go, the only thing it can do is invoke the base class's constructor, and it is the base class's constructor that's responsible for initializing its own members. — Sam Varshavchik
– Sam Varshavchik, Commented Oct 3, 2018 at 16:38
@SamVarshavchik Ah I didn't know that, surprised I haven't run into it before. I'm doing some weird mix-in stuff at the moment so that's why. I'll stick with using this->a = v; — gct
– gct, Commented Oct 3, 2018 at 16:39

NathanOliver · Accepted Answer · 2018-10-03 16:56:53Z

5

You need to initialize the base class in the member initialization list of the derived class. Since your base doesn't have a constructor you can use curly brace initialization (uniform initialization) like

template <typename T>
struct base {
    T a;
};

template <typename T>
struct derived : base<T> {
    derived(T v) : base<T>{v} {}
};

edited Oct 3, 2018 at 16:56

answered Oct 3, 2018 at 16:51

NathanOliver

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1 Comment

NathanOliver Over a year ago

You edited out the comment right as I did. That was pretty quick :)

scohe001 · Accepted Answer · 2018-10-03 16:39:30Z

2

Create a constructor in the base class and call it in the initialization list of the derived class. ie:

template <typename T>
struct base {
    T a;
    base(T v) : a(v) { }
};

template <typename T>
struct derived : base<T> {
    derived(T v) : base(v) {}
};

answered Oct 3, 2018 at 16:39

scohe001

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2 Comments

gct Over a year ago

That's the consensus, unfortunately my number of constructor arguments depends on a template parameter (dimensionality for a vector class), so I was inheriting to provide proper constructors, I'll just use the this-> syntax

scohe001 Over a year ago

@SeanMcAllister you can always overload the base constructor for different cases.

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