Simple Macro Function Pointer Trick...

[Chris Thomasson]

I was wondering if the following technique will produce undefined behavior:
_______________
#include <cstdio>

#define CALL_MACRO_FUNCTION(func_ptr)func_ptr()

#define MY_MESSAGE() "Press <ENTERto exit."

int main(void) {
puts(CALL_MACRO_FUNCTION(MY_MESSAGE));
getchar();
return 0;
}

_______________
I define 'MY_MESSAGE' as a macro function and only pass the name of it to
'CALL_MACRO_FUNCTION' which in turn uses the name to actually invoke it. The
name of a macro function is analogous to a function pointer. This is useful
when you need to have precise control over when a macro will actually
expand.
--
Chris M. Thomasson
http://appcore.home.comcast.net

[Chris Thomasson]

"James Kanze" <ja*********@gmail.comwrote in message
news:11**********************@n39g2000hsh.googlegr oups.com...
On Sep 14, 4:16 pm, "Chris Thomasson" <[e-mail]:

>I was wondering if the following technique will produce undefined
behavior:
_______________
#include <cstdio>

[...]

>_______________

Not related to your actual question, but the above shouldn't
compile with a conformant implementation of <cstdio>.

[...]

DOH! Ah crap. Of course your correct. Here is some better code:

________________
#include <cstdio>

#define CALL_MACRO_FUNCTION(func_ptr)func_ptr()

#define MY_MESSAGE() "Press <ENTERto exit."

int main(void) {
{
using namespace std;
puts(CALL_MACRO_FUNCTION(MY_MESSAGE));
getchar();
}

return 0;
}

________________

>I define 'MY_MESSAGE' as a macro function and only pass the
name of it to 'CALL_MACRO_FUNCTION' which in turn uses the
name to actually invoke it. The name of a macro function is
analogous to a function pointer. This is useful when you need
to have precise control over when a macro will actually
expand.

It should work.

Yeah. I agree. BTW, thank you for you response.