Somewhere in the E.U., le 04/09/2001

 Hi

  According to my (probably flawed) understanding of the standard, the
folowing should output "float". My compiler disagrees. Who's right? A
Google search turned up only irrelevancies.

 ---- 8>< ---------------------- ><8 ----

#include <iostream>


#define TEST(type) \
::std::cout << "##type" << ::std::endl;


main()
{
 TEST(float)
}


#undef TEST

 ---- 8>< ---------------------- ><8 ----

   Hubert Holin
   Hubert.Holin@Bigfoot.com

I am not a mad scientist, I am a mad mathematician!

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Hubert HOLIN wrote:

>
> #define TEST(type) \
> ::std::cout << "##type" << ::std::endl;
>
I think you're confused.  The ## glues two tokens together.
It also has to be between two tokens to be the ## operator.
In this case it's embedded in the middle of a string literal
(which is one token in itself).

What you really want is the # preprocessor operator:

 #define TEST(type) \
  std::cout << #type << std::endl;

#define TEST(float) then maps to

 std::cout << "float" << std::endl;

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