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From: Rick.Regan on 27 Jun 2010 00:11
You said "If you stop in the debugger and examine the value of the double, you will find it to be -937566.23646998685 ... on output
VC++ is rounding whereas G++ is truncating."

The debugger is not printing all the digits. Try printing the double in hex with the %a format specifier ("%.13a" gets all the
digits in VC++). Now take that value, translate it to binary, and then convert it to decimal. It is exactly
-937566.236469986848533153533935546875. G++ is not truncating; it is rounding the exact value to 16 digits (which only looks like
truncating since the remaining digits are less than 1/2 ULP). VC++ is rounding up, which is incorrect.



George Neuner wrote:

I do not have a cite ready, but problems similar to yours have beendiscussed
26-Jun-10

I do not have a cite ready, but problems similar to yours have been
discussed before in this group. The last one I remember was about
exponential notation: someone was reading in values like "0.0003" and
"3.e-4" and finding they do not compare equal. This is a known issue
with the VC++ library - it goes back at least to VS2002 and maybe
further.



Does the file have to be human readable? If not, store the exact bit
pattern in binary or in hexadecimal (16 chars).



I do not deny there is a discrepancy, but the question still is which
compiler is wrong. I see that VC++ is giving back the input value,
but it is by coincidence.

If you stop in the debugger and examine the value of the double,
you will find it to be -937566.23646998685 ... on output VC++ is
rounding whereas G++ is truncating.

If you change the input string to "-937566.2364699868", the value of
the double is still -937566.23646998685 ... but in this case, G++
matches the input while VC++ does not.

For my money VC++ is the one doing it wrong.

George


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