Why can't templates be within extern “C” blocks?

Question

This is a follow-up question to an answer to Is it possible to typedef a pointer-to-extern-“C”-function type within a template?

This code fails to compile w

Answer 1

Because extern C disables name mangling, which templates use

To see that templates are implemented with name mangling, compile and decompile:

#include 

template 
C f(C i) { return i; }

int main() {
    f(1);
    f(1.5);
}

with:

g++ -c -g -std=c++98 main.cpp
objdump -Sr main.o

The output contains:

int main() {
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
   4:   48 83 ec 10             sub    $0x10,%rsp
    f(1);
   8:   bf 01 00 00 00          mov    $0x1,%edi
   d:   e8 00 00 00 00          callq  12 
            e: R_X86_64_PC32    _Z1fIiET_S0_-0x4
    f(1.5);
  12:   48 b8 00 00 00 00 00    movabs $0x3ff8000000000000,%rax
  19:   00 f8 3f 
  1c:   48 89 45 f8             mov    %rax,-0x8(%rbp)
  20:   f2 0f 10 45 f8          movsd  -0x8(%rbp),%xmm0
  25:   e8 00 00 00 00          callq  2a 
            26: R_X86_64_PC32   _Z1fIdET_S0_-0x4
}
  2a:   b8 00 00 00 00          mov    $0x0,%eax
  2f:   c9                      leaveq 
  30:   c3                      retq

Note how all callq were made to call weird names like _Z1fIiET_S0_.

The same goes for other features which depend on name mangling, e.g. function overloading.

I have written a more detailed answer at: What is the effect of extern "C" in C++?