g++ 4.2.1: -O breaks linking to a templated function

Question

g++ main.c f.c below works with g++-4.2.1, but
g++ -O3 main.c f.c gives the warning

/usr/libexec/gcc/powerpc-apple-darwin8/4.2.1/ld: Undefined symbols:
int f<int>(int const*)
collect2: ld returned 1 exit status


// main.c
template <typename T>
int f( const T* A );

int main()
{
    int* A = new int[10];
    int ftemplate = f( A );
}


// f.c
template <typename T>
int f( const T* A )
{   return A[0];
}

int call_f()
{   int* A = new int[10];
    return f( A );  // ok here but not from main()
}

On macosx 10.4.11 powerpc-apple-darwin8-g++-4.2.1 (GCC) 4.2.1 (Apple Inc. build 5564), -O2 works, -O3 does not.
On macosx 10.7.4 i686-apple-darwin11-llvm-g++-4.2 (from https://github.com/kennethreitz/osx-gcc-installer),
plain g++ *.c works, g++ -O *.c gives the same ld: Undefined symbols error.
Maybe a bug g++ <-> old /usr/bin/ld ? More likely I've done something stupid ...

Can anyone help, or see if this works on a non-Mac ? Thanks !

@H2CO3, ok, but that's not the question: on 10.7, -O2 doesn't work either. And why should it break the linker ? — denis, Nov 20 '12 at 18:35

score 1 · Accepted Answer · answered Nov 20 '12 at 20:08

1

Unless you explicitly instantiate a function template for the arguments you use in a function call, you have to make the function template definition visible to the caller of it.

This includes the call in main.

It probably works in unoptimized builds because the compiler emits an exported function definition symbol for the implicit function template instantiation. The C++ Standard grants compilers to omit doing that, and GCC does it here for optimized builds (probably it just inlines the call and then the definitions symbol becomes unused).

answered Nov 20 '12 at 20:08

Johannes Schaub - litb

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Thanks -- my misunderstanding, duplicate of [template-programming-in-c](http://stackoverflow.com/questions/3765556/template-programming-in-c) on SO, but confusing because it works sometimes. – denis Nov 21 '12 at 17:27
it is not wrong always. you can limit the types usable and instantiate the template explicitly for those types. – Johannes Schaub - litb Nov 21 '12 at 17:34
can you tell me more: how can I say e.g. template < T float or double only > ? (in g++ today, not c++11) – denis Nov 22 '12 at 17:01
you best ask this as a new stackoverflow question – Johannes Schaub - litb Nov 22 '12 at 20:16

g++ 4.2.1: -O breaks linking to a templated function

1 Answers1