C/C++ counting the number of decimals?

Question

Lets say that input from the user is a decimal number, ex. 5.2155 (having 4 decimal digits). It can be stored freely (int,double) etc.

Is there any clever (or very simple) way to find out how many decimals the number has? (kinda like the question how do you find that a number is even or odd by masking last bit).

Answer 1

Two ways I know of, neither very clever unfortunately but this is more a limitation of the environment rather than me :-)

The first is to sprintf the number to a big buffer with a "%.50f" format string, strip off the trailing zeros then count the characters after the decimal point. This will be limited by the printf family itself. Or you could use the string as input by the user (rather than sprintfing a floating point value), so as to avoid floating point problems altogether.

The second is to subtract the integer portion then iteratively multiply by 10 and again subtract the integer portion until you get zero. This is limited by the limits of computer representation of floating point numbers - at each stage you may get the problem of a number that cannot be represented exactly (so .2155 may actually be .215499999998). Something like the following (untested, except in my head, which is about on par with a COMX-35):

count = 0
num = abs(num)
num = num - int(num)
while num != 0:
    num = num * 10
    count = count + 1
    num = num - int(num)

If you know the sort of numbers you'll get (e.g., they'll all be 0 to 4 digits after the decimal point), you can use standard floating point "tricks" to do it properly. For example, instead of:

while num != 0:

use

while abs(num) >= 0.0000001:

Answer 2

Once the number is converted from the user representation (string, OCR-ed gif file, whatever) into a floating point number, you are not dealing with the same number necessarily. So the strict, not very useful answer is "No".

If (case A) you can avoid converting the number from the string representation, the problem becomes much easier, you only need to count the digits after the decimal point and subtract the number of trailing zeros.

If you cannot do it (case B), then you need to make an assumption about the maximum number of decimals, convert the number back into string representation and round it to this maximum number using the round-to-even method. For example, if the user supplies 1.1 which gets represented as 1.09999999999999 (hypothetically), converting it back to string yields, guess what, "1.09999999999999". Rounding this number to, say, four decimal points gives you "1.1000". Now it's back to case A.