[SOLVED] Can we see the templates instantiated by the C++ compiler?

Can we see the templates instantiated by the C++ compiler?

Is there a way to see the compiler-instantiated code for a function template or a class template in C++?

Assume I have the following piece of code:

template <class T> T add(T a, T b) {
    return a + b;
}

When I call:

add<int>(10, 2);

... I would like to see the function that the compiler creates for the int template specialization.

I am using g++, VC++, and need to know the compiler options to achieve this.

Solution

You can definitely see the assembly code generated by the g++ using the "-S" option.

I don't think it is possible to display the "C++" equivalent template code - but I would still want a g++ developer to chime in why - I don't know the architecture of gcc.

When using assembly, you can review the resulting code looking for what resembles your function. As a result of running gcc -S -O1 {yourcode.cpp}, I got this (AMD64, gcc 4.4.4)

_Z3addIiET_S0_S0_:
.LFB2:
    .cfi_startproc
    .cfi_personality 0x3,__gxx_personality_v0
    leal    (%rsi,%rdi), %eax
    ret
    .cfi_endproc

Which really is just an int addition (leal).

Now, how to decode the c++ name mangler? there is a utility called c++filt, you paste the canonical (C-equivalent) name and you get the demangled c++ equivalent

qdot@nightfly /dev/shm $ c++filt 
_Z3addIiET_S0_S0_ 
int add<int>(int, int)