This code works (C# 3)
double d;
if(d == (double)(int)d) ...;
A simple test such as 'x == floor(x)' is mathematically assured to work correctly, for any fixed-precision FP number.
All legal fixed-precision FP encodings represent distinct real numbers, and so for every integer x, there is at most one fixed-precision FP encoding that matches it exactly.
Therefore, for every integer x that CAN be represented in such way, we have x == floor(x) necessarily, since floor(x) by definition returns the largest FP number y such that y <= x and y represents an integer; so floor(x) must return x.