Count number of digits after `.` in floating point numbers?
c, floating-point
Solution
The problem isn't really solvable as stated, since floating-point is typically represented in binary, not in decimal. As you say, many (in fact most) decimal numbers are not exactly representable in floating-point.
On the other hand, all numbers that are exactly representable in binary floating-point are decimals with a finite number of digits -- but that's not particularly useful if you want a result of 2 for `3.44`.
When I run your code snippet, it says that `3.44` has 2 digits after the decimal point -- because `3.44 * 10.0 * 10.0` just happens to yield exactly `344.0`. That might not happen for another number like, say, `3.43` (I haven't tried it).
When I try it with `1.0/3.0`, it goes into an infinite loop. Adding some `printf`s shows that `no` becomes exactly `33333333333333324.0` after 17 iterations -- but that number is too big to be represented as an `int` (at least on my system), and converting it to `int` has undefined behavior.
And for large numbers, repeatedly multiplying by 10 will inevitably give you a floating-point overflow. There are ways to avoid that, but they don't solve the other problems.
If you store the value `3.44` in a `double` object, the actual value stored (at least on my system) is exactly `3.439999999999999946709294817992486059665679931640625`, which has 51 decimal digits in its fractional part. Suppose you really want to compute the number of decimal digits after the point in `3.439999999999999946709294817992486059665679931640625`. Since `3.44` and `3.439999999999999946709294817992486059665679931640625` are effectively the same number, there's no way for any C function to distinguish between them and know whether it should return 2 or 51 (or 50 if you meant `3.43999999999999994670929481799248605966567993164062`, or ...).
You could probably detect that the stored value is "close enough" to `3.44`, but that makes it a much more complex problem -- and it loses the ability to determine the number of decimal digits in the fractional part of `3.439999999999999946709294817992486059665679931640625`.
The question is meaningful only if the number you're given is stored in some format that can actually represent decimal fractions (such as a string), or if you add some complex requirement for determining which decimal fraction a given binary approximation is meant to represent.
There's probably a reasonable way to do the latter by looking for the unique decimal fraction whose nearest approximation in the given floating-point type is the given binary floating-point number.
Problem
This is one interview question. How do you compute the number of digit after `.` in floating point number. e.g. if given 3.554 output=3 for 43.000 output=0. My code snippet is here ``` double no =3.44; int count =0; while(no!=((int)no)) { count++; no=no*10; } printf("%d",count); ``` There are some numbers that can not be indicated by `float` type. for example, there is no `73.487` in `float` type, the number indicated by `float` in c is `73.486999999999995` to approximate it. Now how to solve it as it is going in some infinite loop. Note : In the IEEE 754 Specifications, a 32 bit float is divided as 24+7+1 bits. The 7 bits indicate the mantissa.