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

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

now when i call

add<int>(10,2); 

I would like to know the f开发者_运维知识库unction that compiler creates for int specific version.

I am using G++, VC++. It will be helpful if some can help me point out the compiler options to achieve this.

Hope the question is clear. Thanks in advance.

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);
}

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)

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)

Now there is an on-line tool which does this for you: https://cppinsights.io/ For example, this code

template<class X, class Y> auto add(X x, Y y) {
  return x + y;
}

int main()
{
  return add(10, 2.5);
}

Is translated to

template<class X, class Y> auto add(X x, Y y) {
  return x + y;
}

/* First instantiated from: insights.cpp:9 */
#ifdef INSIGHTS_USE_TEMPLATE
template<>
double add<int, double>(int x, double y)
{
  return static_cast<double>(x) + y;
}
#endif


int main()
{
  return static_cast<int>(add(10, 2.5));
}

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.

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

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.

I think c++ Insights is what you want.