P.P.S. Never mind, passing f = 0.12345678901234567 is a counterexample. So
the answer is no, you won't get a bitwise equivalent if you do a
stringValue-floatValue round trip.
--Andy
On Dec 12, 2009, at 1:25 PM, Andy Lee wrote:
> On Dec 12, 2009, at 1:07 PM, Andy Lee wrote:
>> If I understand the question, it's not about converting an arbitrary decimal
>> string to a float, but specifically a string that was generated from a float
>> in the first place.
>>
>> As glenn pointed out, that string most certainly *can* be a string that can
>> be converted back to a bitwise equivalent of the original float. Ben's
>> question was whether in practice it is guaranteed to be so, or whether
>> stringValue uses a maximum number of decimal places that would lead to
>> rounding error. I can't tell from a quick look at the docs.
>
> P.S. If the answer is no, it seems to me there should be an easy
> counterexample, but my math isn't good enough to figure one out. I've tried
> the following code with f = 0.1, 0.100001, and 0.12345, but I always get back
> a bitwise match for the original float, even when stringValue returns a
> rounded string.
>
> - (void)testFloat:(float)f
> {
> NSString * s = [[NSNumber numberWithFloat:f] stringValue];
> float f2 = [s floatValue];
> NSLog(@"%.20f, %@, %.20f", f, s, f2);
> char *cp1 = (char *)(&f);
> char *cp2 = (char *)(&f2);
> int i;
> for (i = 0; i < sizeof(float); i++)
> {
> NSLog(@"byte %d %@", i, (cp1[i] == cp2[i] ? @"MATCHES" : @"DOES NOT
> match"));
> }
> }
>
> --Andy
>
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