Error message "undefined reference to template function passed as template parameter"

Question

When I pass a template function as a template parameter of a base class, the linker complains that it cannot link the function:

#include <stdio.h> template<int I> inline int identity() {return I;} //template<> inline int identity<10>() {return 20;} template<int (*fn)()> class Base { public: int f() { return fn(); } }; template<int Val> class Derived : public Base<identity<10> > { public: int f2() { return f(); } }; int main(int argc, char **argv) { Derived<10> o; printf("result: %d\n", o.f2()); return 0; }

Results in:

$ g++ -o test2 test2.cpp && ./test2 /tmp/ccahIuzY.o: In function `Base<&(int identity<10>())>::f()': test2.cpp:(.text._ZN4BaseIXadL_Z8identityILi10EEivEEE1fEv[_ZN4BaseIXadL_Z8identityILi10EEivEEE1fEv]+0xd): undefined reference to `int identity<10>()' collect2: error: ld returned 1 exit status

If I comment out the specialization, then the code compiles and links as expected. Also, if I inherit from Base<identity<Val> > instead of Base<identity<10> >, the code works as I expect.

Try here: http://coliru.stacked-crooked.com/a/9fd1c3aae847aaf7

What do I miss?

This problem seems to be a gcc bug: it compiles and links OK using clang and icc. BTW, the name identity() is normally used for transformation where the result is identical to the argument. — Dietmar Kühl
– Dietmar Kühl, Commented Dec 30, 2016 at 16:04
The workaround: class Derived : public Base<static_cast<int(*)()>(identity<10>) >. live demo — W.F.
– W.F., Commented Dec 30, 2016 at 16:07
@melpomene: sure. However, the template parameter seems to be more something like an index (as in f<sub>i</sub>()) than a function argument. — Dietmar Kühl
– Dietmar Kühl, Commented Dec 30, 2016 at 16:07

Dietmar Kühl · Accepted Answer · 2016-12-30 16:16:03Z

It seems the problem is a gcc error: the code compiles and links with clang, icc, and the EDG frontend. A potential work-around not changing any of the uses would be the use of a class template identity instead of a function:

template<int I> struct identity { operator int() { return I; } }; template<typename fn> class Base { public: int f() { return fn(); } };

Not making the function template inline could also work. Isn't linker supposed to remove duplicate template instantiations anyway?

Peter Mortensen · Accepted Answer · 2017-01-03 21:16:35Z

Lifting it out into a typedef makes it compile, i.e.

typedef Base< identity<10> > base10;

I am not quite sure why doing it straight in the class definition doesn't work.

http://coliru.stacked-crooked.com/a/f00b4f4d1c43c2b0

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