Can we see the template instantiated code by C++ compiler

Question

is there a way to know the compiler instantiated code for a template function or a class in C++

Assume I have the following piece of code

template &l         


        

Answer 1

When the optimizer has done its deeds, you most likely have nothing left that looks like a function call. In your specific example, you'll definitely end up with an inlined addition, at worse. Other than that, you can always emit the generated assembler in a separate file during compilation, and there lies your answer.

Answer 2

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)

Answer 3

If you want to see the assembly output, use this:

g++ -S file.cpp

If you want to see some (pseudo) C++ code that GCC generates, you can use this:

g++ -fdump-tree-original file.cpp

For your add function, this will output something like

;; Function T add(const T&, const T&) [with T = int] (null)
;; enabled by -tree-original

return <retval> = (int) *l + (int) *r;

(I passed the parameters by reference to make the output a little more interesting)

Answer 4

The easiest is to inspect the generated assembly. You can get an assembly source by using -S flag for g++.

Answer 5

Clang (https://clang.llvm.org/) can pretty-print AST of instantiated template:

For your example:

test.cpp

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

void tmp() {
    add<int>(10,2); 
}

Command to pretty-print AST:

$ clang++ -Xclang -ast-print -fsyntax-only test.cpp

Clang-5.0 output:

template <class T> T add(T a, T b) {
    return a + b;
}
template<> int add<int>(int a, int b) {
    return a + b;
}
void tmp() {
    add<int>(10, 2);
}

Answer 6

If your looking for the equivalent C++ code then no. The compiler never generates it. It's much faster for the compiler to generate it's intermediate representation straight off than to generate c++ first.