问题
Entering a value such as 27.8675309 into the "Decimal representation" field of the IEEE 754 Converter changes the value I entered to 27.86753. Likewise, Java drops the last two digits when a parse a string with the same value.
Float.parseFloat("27.8675309") // Results in a float value of 27.86753
I am not sure what the "Decimal representation" of the IEEE converter actually is (is it a float?) but I would expect it to give me the biggest number possible that:
- Is a float value
- Does not exceed the original value I entered
I would expect Java to behave in a similar fashion, that is, I would expect the line of code above to return a float value equal to 27.8675308 or an even larger float value that is closer to my original input instead of just dropping decimal places. What am I missing here?
回答1:
This is as expected.
If you look at the binary representation of 27.8675309 (27.867530822753906 as a double):
01000001110111101111000010110100
the next highest value is:
01000001110111101111000010110101
which yields 27.867533 (27.86753273010254 as a double), and the next lowest value is:
01000001110111101111000010110011
which yields 27.867529 (27.867528915405273 as a double)
There simply aren't enough bits in the mantissa of a Float to represent any value in between, so your value is rounded down in decimal to 27.867530
回答2:
As Alnitak mentioned, you've run out of bits. You could use Double for higher precision, though. Double implements 64-bit IEEE 754 floating point whereas Float implements 32-bit IEEE 754 floating point.
来源：https://stackoverflow.com/questions/9083905/why-is-my-float-being-truncated