Raymond Hettinger <[email protected]> added the comment:
The suggested examples are misleading because they use 6.24 which is not
exactly representable in binary floating point. Representation issues are
orthogonal to the OP's issue which is really just a simple rounding example:
>>> x = float.fromhex('0x0.1234560000001')
>>> unpack('!f', pack('!f', x))[0].hex()
'0x1.2345600000000p-4'
Also, if something like the suggested note is adopted, it needs to be worded in
a way that doesn't imply that the struct implementation is broken or
misdesigned.
A better note would focus on the basic (and obvious) fact that downgrading from
double precision to single precision
entails a loss of precision.
----------
nosy: +rhettinger
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