is there a possibility that casting a double created via Math.round() will still result in a truncated down number No, round() will always round your double to the correct value, and then, it will be cast to an long which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining. Here are the docs from Math.round(double): Returns the closest long to ...
Casting can be used to clearly state that you are calling a child method and not a parent method. So in this case it's always a downcast or more correctly, a narrowing conversion.
Essentially casting will not change anything in how it works, it does exactly what it says, allocates memory, and casting does not effect it, you get the same memory, and even if you cast it to something else by mistake (and somehow evade compiler errors) C will access it the same way. Edit: Casting has a certain point.
Static cast is also used to cast pointers to related types, for example casting void* to the appropriate type. dynamic_cast Dynamic cast is used to convert pointers and references at run-time, generally for the purpose of casting a pointer or reference up or down an inheritance chain (inheritance hierarchy). dynamic_cast (expression)
Casting a variable expression to void to suppress this warning has become an idiom in the C and later C++ community instead because the result cannot be used in any way (other than e.g. (int)x), so it's unlikely that the corresponding code is just missing.
c++ - What does casting to void really do? - Stack Overflow