Question

I would like to be able to write a macro CONDITIONALFUNCTION so that

CONDITIONALFUNCTION( FunctionName )
{
    ConditionalExpression()
}

expands to

bool FunctionName( const Arguments& args )
{
    return ConditionalExpression();
}

Is this even possible?

The closest I can find on SO is this thread:

Possible to define a function-like macro with a variable body?

except unlike in that thread, I have the additional requirement that the “body” within the braces is not a complete valid C++ statement, but rather an expression to be wrapped (effectively) in an ‘if’ statement.

Please assume I already know this may be impossible, and is almost surely stupid and evil 🙂

Answer 1

I’m going to assume you’ve got a good reason for using macros in the first place…

It’s not possible with the syntax you’ve given with the question.

The closest workable macro syntax is:

#define CONDITIONALEXPRESSION(f, c) \
bool f( const Arguments& args ) \
{ return c; }

CONDITIONALEXPRESSION(FunctionName, ConditionalExpression())

This will expand to the same as the expanded function in the question