如何避免构造函数中的冗余计算

Question

I've got a constructor like so:

C(T x) : base(f(x))
{
   ...
   do something with f(x)
   ...
}

f(x) is not exposed as a member in the base class. How do I avoid calculating f(x) twice if I can't modify C s base?

Answer 1

You could use two constructors, for example:

private C(WhateverFReturns x) : base(x)
{
    //do something with x
}

public C(T x) : this(f(x))
{

}