What this expression is doing is it first declares the existence of a long called x, and then assigning it the value of the right hand side expression. The right hand side expression is 1/2, and since 1 and 2 are both integers this is interpreted as integer division. With integer division the result is always an Integer, so something along the lines of 5/3 will return 1, as only one three fits in a five. So with 1/2, how many 2s can fit into 1? 0.
This can in some languages result in some interesting outputs if you write something like
double x = 1/2. You might expect 0.5 in this case, but it will often evaluate the integer value on the right first before assigning and converting the result into a double, giving the value 0.0
It is important to note that when doing this kind of type conversion, it will never round the result. So if you do the opposite:
long x = (long)(1.0/2.0);
then while (1.0/2.0) will evaluate to 0.5, the (long) cast will force this to be truncated to 0. Even if I had long x = (long)(0.9), the result will still be 0. It simply truncates after the decimal point.
