Failed to specialize function template compiler error

Question

I have a template vector class, and I am trying to overload the arithmetic operators such that adding/subtracting vectors of different primitive types will yield a vector of the primitive type returned by the arithmetic operation (i.e., adding a vector of int and a vector of double will yield a vector of double). The relevant code follows:

template<typename T> class Vector { private: T x, y, z; public: // constructors, destructors, etc. T X() const { return ( this->x ); } T Y() const { return ( this->y ); } T Z() const { return ( this->z ); } template<typename U> auto operator+(const Vector<U> &v) const -> Vector<decltype(this->x + v.X())> { return ( Vector<decltype(this->x + v.X())>( this->x + v.X(), this->y + v.Y(), this->z + v.Z() ) ); } };

Then, in main.cpp:

Vector<double> d1(1,2,3), d2(4,5,6); std::cout << d1 + d2;

I get the following error from Visual Studio 2012:

error C2893: Failed to specialize function template 'Vector<T> Vector<double>::operator +(const Vector<U> &) const' with [ T=unknown ] With the following template arguments: 'double' error C2676: binary '+' : 'Vector<T>' does not define this operator or a conversion to a type acceptable to the predefined operator with [ T=double ]

Please advise.

Try to upgrade to visual-studio-2013. works with gcc (ideone.com/GlJOcE) — Jarod42
– Jarod42, Commented Jan 27, 2015 at 16:58

danielschemmel · Accepted Answer · 2015-01-27 17:13:35Z

1

Visual C++ fails to determine the correct type of your decltype expression. By replacing your expression with decltype(*((T*)0) + *((U*)0)) you remove the dependency on this and the argument v, which leads to the intended result on VS as well.

Note: Bug was reproducable on VS2013.4 on which the fix was tested as well.

answered Jan 27, 2015 at 17:13

danielschemmel

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5 Comments

LRC Over a year ago

Thank you! That works. I am confused by the expression you have here; can you explain its meaning?

danielschemmel Over a year ago

@LRC It dereferences a (null) pointer of appropriate type to create an lvalue of T and U respectively. This way, they need not be constructible in any specific way.

LRC Over a year ago

Thank you for the explanation! That makes perfect sense!

Jarod42 Over a year ago

decltype(std::declval<T>() + std::declval<U>()) seems more readable.

danielschemmel Over a year ago

@Jarod42 I agree, I was just too lazy to figure out if declval was supported on VS2012 already.

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