Re: Float comparison

Richard Heathfield wrote:
CBFalconer said:
Keith Thompson wrote:
Joe Wright <joewwright@xxxxxxxxxxx> writes:
Keith Thompson wrote:
Joe Wright <joewwright@xxxxxxxxxxx> writes:

[...]
float x = 0.3333333432674407958984375;
float y = 3.33333343e-01;

... snip ...

The approximation 0.333333343, though it's perfectly fine in
most contexts, doesn't demonstrate this.

Yes it does. Both the value above are members of the same range,

There are infinitely many ranges of which both values are members.

which is the range specified by the floating object when
initialized with either value.

No, the floating point object doesn't specify a range. It is able
to store a value, but only one value at a time. We covered this
before.

No. Maybe you misstated it before. The point is that a given FP
system always specifies a range covered by a particular FP value.
The mere fact that you say this indicates you haven't a clue as to
what is really going on.


Which is also why (x == y) is evaluated as true.

It's evaluated as true when x has the same value as y.

The only difference is in the programming which initialized them.

The programming which initialized them doesn't matter as far as
their value is concerned. The thing that affects their value is,
would you believe, the last value to be stored there. If that's the
same for both, both will store the same value. In this case, the
value stored in y is the result of converting a double-precision
floating-point value such as (on my system)
0.333333343000000004163752009844756685197353363037109375 into a
single-precision floating point value.

You just attempted to specify the programming. Wrong. The only
thing available is the FP object itself.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
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