CODEWITHSUNDEEP
c++c++11macrosoperators
George
George

Reputation: 52

C++ Macros and overloading

I am working with c++ Macros, trying to implement on certain pattern. I have the following 3 macros:

#define First Objct t; t
#define Second a() / b()
#define Third ;

I've overload the lambda-ops in format: Objct [ Objct ]

So when I have in main() a line like: 

First [ Second ] Third

it works fine, as it's 'translated' into: 

Objct t; t[a()/b()]

(note: a() and b() are dummy functions returning type object.)

The hard part, is that I also have to make it work without the lambdas.

First Second Third

which means 

Objct t; t a() / b()

In that case, I got a semicolon missing from FIRST to SECOND. I'm trying to figure out, what changes could be made (probably) to FIRST macro so it can compile in both cases. I'm not sure if I managed to explain myself properly. Any ideas?

Upvotes: 0

Views: 100

Answers (1)

Christophe
Christophe

Reputation: 73456

Your question in pretty strange. I would strongly advise you not to use such a weird construct in real code. But I've looked at your question as if it were a funny puzzle.

Potential solution

I think there is no way to make both statement compile by changing only First. But if you change Second to: 

#define Second +0,a() / b()

it compiles in both cases, under the sole condition that operator + is defined for Objct in combination with an int. If you manage to implement this operator without side effect, it would even produce what you expect.

Live demo

Why does it work ?

This definition makes use of the coma operator, the only issue being that the coma operator requires two expressions. +0 solves the issue syntactically, as +0 alone is valid, and t+0 is valid with the above-mentionned requirement.

With such a definition, First Second Third is preprocessed as 

 Objct t; t +0,a() / b() ;     // comma operator evaluates t+0  
                               // then a()/b()

And First [Second] Third would be preprocessed as 

 Objct t; t [+0,a() / b()] ;  // comma operator makes +0 being evaluated
                              // and lost and [] is called with value of a()/b().

Upvotes: 1

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