How to check if float is a whole number?
c++, decimal, floating-point, rounding
Solution
There isn't really a simple answer
Integral values do have exact representations in the float and double formats. So, if it's really already integral, you can use:
f == floor(f)
However, if your value is the result of a calculation which at one point involved any sort of non-zero fractional part, then you will need to be concerned that you may have something very close to an integer but which isn't really, exactly, to-the-last-bit the same. You probably want to consider that to be integral.
One way this might be done:
fabs(f - round(f)) < 0.000001
And while we are on the subject, for the purists, we should note that `int i = f;` or `double i = f;` will round according to the FPU mode whereas round(3) will round half-way cases away from zero.
Problem
The `printf` function's `%g` is able to show the whole number `3` if the float is 3.00, and will show `3.01` if the float's value isn't a round number float. How would you do this yourself through some code, without formatting the number as a string?