Re: Function Pop-Up and #if 0
Re: Function Pop-Up and #if 0
- Subject: Re: Function Pop-Up and #if 0
- From: Bryan Pietrzak <email@hidden>
- Date: Fri, 9 Jul 2004 12:56:52 -0500
Sure, that makes sense. But since "#if 0" is a pretty common thing that
a lot of people use, do you expect to make the scanner just a tad bit
more knowledge about the preprocessor in this one respect.
Bryan
On Jul 9, 2004, at 10:54 AM, David Ewing wrote:
The scanner that is used for the function popup is intentionally
simple so that it can be tolerant of syntax errors. One side effect of
this is that only has limited knowledge of the preprocessor. This is a
simple example of the kind of stuff that confuses it.
Dave
On Jul 8, 2004, at 3:52 PM, Bryan Pietrzak wrote:
I just filed rdar://3722111
This code will cause the function pop-up to stop showing any symbol
after function two....
--------------------------------------------------
void one(void)
{
}
void two(void)
{
#if 0
}
#endif
return err;
}
void three(void)
{
}
--------------------------------------------------
Shouldn't everything inside #if 0 ... #endif be ignored?
Bryan
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