问题
Is there any efficient way (without converting the float into a string) to obtain the number of digits a floating-point number consists of (independent of its length and precision) ?
On that way I can implement a fairly good, portable, problematic-less function for comparison/conditioning by multiplying the float by the number of the digits it consists of.
回答1:
Q: Is there any efficient way to obtain the number of digits a floating-point number?
A: I doubt it.
Every finite FP number is exact, but maybe not the exact value one thinks.
Due to typical binary64 implementation of a double,double x = 0.53, x value: .5300000000000000266453525910037569701671600341796875, 52 digits.double x = 0.1, x value:.1000000000000000055511151231257827021181583404541015625 55 digits.
The next closest double to mathematical 0.1 is .09999999999999999167332731531132594682276248931884765625 with 56 digits.
DBL_MAX: in decimal, typically about 300 digits 17976931348623158... ending with 6728515625..
DBL_MIN: typically 0.000000(~300 zeros) 22250738585072014... (maybe about 700 more digits).
Comparison of FP numbers need not determine the number of digitis in its decimal representation. To compare FP numbers, use the usual relationship operators >, >=, ==, etc.
Theses values are illustrative. YMMV.
来源：https://stackoverflow.com/questions/31690417/count-the-number-of-digits-of-a-floating-point-number