Re: 12.34f vs (float) 12.34

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From: Jon Shemitz (jon_at_midnightbeach.com)
Date: 02/12/04 

Date: Thu, 12 Feb 2004 13:31:46 -0800

Stu Smith wrote:

> > * Very straightforwardly, whereby "(float) 1.23" constructs a double
> > value 1.23, then casts it to a single.

> As far as I'm aware (and we're at the limits of my knowledge here), constant
> folding is done at compile time, so option one is out of the question.

We're pretty much at the limits of my knowledge, too. I've written a
simple compiler, so I have some knowledge of the issues involved, but
that's about it.

But I don't see why you say that option 1 can be ruled out? Seems to
me: Parser sees a stream of digits followed by a . and another stream
of digits which is NOT followed by one of {d, f, m}. Bingo: a double
constant for the parse tree. Optimizer later comes along and sees the
expression cast(float, double constant) and does a compile-time cast,
reducing it to a float constant.

 * * *

I suppose I should just do the brute force thing: write a simple test
program to generate reams of float constants, and see if (say)
2.22222222222222f != (float) 2.22222222222222.

Course, I'll probably find more than I care to know about method size
limits .... 

-- 
programmer, author  http://www.midnightbeach.com
and father          http://www.midnightbeach.com/hs

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