How to get variable name from template g++

Question

How can i get/print variable name from this? I am using arduino Stream to print to console.

#ifndef any_h
#define any_h

#if ARDUINO >= 100
 #include "Arduino.h"
#else
 #include "WProgram.h"
#endif

struct any {
  any(Stream& s):serial(s){}
  template <class T>
  void print(const T& msg)
  {
    getName(class T);
    serial.print(msg);
  }

  template <class A, class... B>
  void print(A head, B... tail)
  {
    print('{');
    print(head);
    print(tail...);
  }

  private:
    Stream& serial; 

};

#endif

Usage:

any A(Serial);

int myInt =34;
float myFloat = 944.5555f;
String myString = " this string";
A.print(myInt,myFloat,myString);

current output

34944.555 this string

I am trying to get something like with the same usage/access or like in this: Demo.

{"variableName":value,"variableName":value}
// That is in this case:
{myInt:34,myFloat:944.55,myString: this string}

What i have already tried:

#define getName(x) serial.print(#x)
void print(const T& msg)
{  
   getName(msg);
   //getName(class T);
   serial.print(msg);
}
output : msg34msg944.555msg this string

C++ does not have reflection. Trying to print the name of the variable without hard coding it is going to be extremely difficult if not impossible. — NathanOliver, Apr 05 '16 at 16:45
@Nigel Without doubt, you have to make a macro that expands to a function call, not a function that uses a macro. Try a [variadic macro](http://stackoverflow.com/questions/679979/how-to-make-a-variadic-macro-variable-number-of-arguments). — LogicStuff, Apr 05 '16 at 16:54
So, just put the function inside the struct... [Demo](http://coliru.stacked-crooked.com/a/8193f7524c17f2f4) (Care that macro doesn't have scope). — Jarod42, Apr 05 '16 at 17:15
@NathanOliver excuse my ignorance but what you mean is in c++ i cannot get the parameter names by pointing to them somehow or looking back at the parameters, is this correct? — Nigel Tiany, Apr 05 '16 at 17:58
@NigelTiany I am just saying that since C++ lacks reflection this becomes a lot harder to do then in a language that has refelection — NathanOliver, Apr 05 '16 at 18:05
@Jarod42 is [this](https://codeshare.io/vYAKB) what you mean? — Nigel Tiany, Apr 05 '16 at 18:10

score 2 · Answer 1 · answered Apr 05 '16 at 19:47

You can use macro to do that combined with stringify preprocessor:

Code:

#include <stdio.h>

#define PRINT_NAME(name) print_name(#name, (name))

void print_name(char *name, char* value) {
    printf("name: %s ---> value: %s\n", name, value);
}

int main (int argc, char* argv[]) {
    char* var1 = 'my var 1';
    char* var2 = 'my new var 2';

    PRINT_NAME(var1);
    PRINT_NAME(var2);

    return 0;
}

Output:

name: var1 --> value: my var 1
name: var2 --> value: my new var 2

I hope this helps :)

Max Lybbert · Answer 2 · 2016-04-08T14:44:11.940

C++ doesn't have reflection, so you're limited to using macros, such as:

#include <iostream>
#define STRINGIFY_IMPL(X) #X
#define STRINGIFY(X) STRINGIFY_IMPL(X)
#define VARIABLE_NAME(X) STRINGIFY(X)
#define VARIABLE_VALUE(X) X

int main()
{
    double d = 3.141;
    std::cout << VARIABLE_NAME(d) << ": " << VARIABLE_VALUE(d) << '\n';
    return 0;
}

(VARIABLE_NAME and VARIABLE_VALUE aren't truly needed, you could also use std::cout << STRINGIFY(d) << ": " << d << '\n';).

You can, of course, combine this all into a super macro (with macros from above):

#include <iostream>
#include <string>
#define STREAM_NAME_VALUE(str, X) str << '{' << STRINGIFY(X) << ':' << X << '}'

int main()
{
    int myInt =34;
    float myFloat = 944.5555f;
    std::string myString = " this string";

    STREAM_NAME_VALUE(std::cout, myInt);
    std::cout << '\n';
    STREAM_NAME_VALUE(std::cout, myFloat);
    std::cout << '\n';
    STREAM_NAME_VALUE(std::cout, myString);
    std::cout << '\n';
    return 0;
}

Not quite. First, macros don' respect scope (they stay defined after the `struct`). Second, if you use these macros inside a function, the information you're looking for (the variable name) will be long gone. The macro will think that all variables are named `msg`, because, at the point the macro is called, all variables are. You need to replace the function with a macro so that the information you're interested in is still around. — Max Lybbert, Apr 06 '16 at 10:08

How to get variable name from template g++

2 Answers2

Code:

Output: