IS operator behaving a bit strangely
- by flockofcode
1) According to my book, IS operator can check whether
expression E (E is type) can be converted to the target type only if E is either a reference conversion, boxing or unboxing. Since in the following example IS doesn’t check for either of the three types of conversion, the code shouldn’t work, but it does:
int i=100;
if (i is long) //returns true, indicating that conversion is possible
l = i;
2)
a)
B b;
A a = new A();
if (a is B)
b = (B)a;
int i = b.l;
class A { public int l = 100; }
class B:A { }
The above code always causes compile time error “Use of unassigned variable”. If condition a is B evaluates to false, then b won’t be assigned a value, but if condition is true, then it will. And thus by allowing such a code compiler would have no way of knowing whether the usage of b in code following the if statement is valid or not ( due to not knowing whether a is b evaluates to true or false) , but why should it know that? Intsead why couldn’t runtime handle this?
b) But if instead we’re dealing with non reference types, then compiler doesn’t complain, even though the code is identical.Why?
int i = 100;
long l;
if (i is long)
l = i;
thank you
