C++11 Optional Template Type Parameter?

Question

If I want to write a class that has an optional type parameter I can do the following:

template<typename T> struct X { T t; }; template<> struct X<void> { }; int main() { X<int> a; X<void> b; };

Is there a way to write it so that the void is unnecessary? ie:

int main() { X<int> a; X b; };

I tried this:

template<typename T = void> struct X { T t; }; template<> struct X<void> { }; int main() { X<int> a; X b; };

but I get:

test.cpp: In function ‘int main()’: test.cpp:16:4: error: missing template arguments before ‘b’ test.cpp:16:4: error: expected ‘;’ before ‘b’

The only solution I know to not using templates is to make a non-templated sub-class. — Some programmer dude
– Some programmer dude, Commented Dec 24, 2012 at 3:03
You did everything correctly in your second attempt, except that you are still required to specify empty <> when instantiating your template with default argument, i.e. it has to be X<> b; — AnT stands with Russia
– AnT stands with Russia, Commented Dec 24, 2012 at 3:24

Mikael Persson · Accepted Answer · 2012-12-24 03:18:03Z

4

You technically need to write:

X<> b;

But you can simply fix that ugliness with a typedef:

typedef X<> Y;

And then you can do:

Y b;

answered Dec 24, 2012 at 3:18

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1 Comment

In which case you can just as well do typedef X<void> Y; and avoid the default value.

David G · Accepted Answer · 2012-12-27 15:25:27Z

1

If only this was possible:

template <typename T> using X = std::is_void<T>::value ? _X<> : _X<T>

But this doesn't compile so unfortunately you are stuck with a typedef like the other answer.

answered Dec 27, 2012 at 1:09

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There is typename std::conditional<std::is_void<T>::value, _X<> : _X<T>>::type. I know this is old, but it's worth sharing.