Is "long x = 1/2" equal to 1 or 0, and why? [duplicate]
if I have something like:
long x = 1/2;
shouldn't this be rounded up to 1? When I print it on the screen it say 0.
Solution 1:
It's doing integer division, which truncates everything to the right of the decimal point.
Solution 2:
Integer division has its roots in number theory. When you do 1/2 you are asking how many times does 2 equal 1? The answer is never, so the equation becomes 0*2 + 1 = 1, where 0 is the quotient (what you get from 1/2) and 1 is the remainder (what you get from 1%2).
It is right to point out that % is not a true modulus in the mathematical sense but always a remainder from division. There is a difference when you are dealing with negative integers.
Hope that helps.
Solution 3:
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.
Solution 4:
It can't round because it's never in a state to be rounded
The expression "1/2" is never 0.5 before assign to long
Now, long x = 1.0/2.0 because the expression on the right before assign is valid for rounding. Unless you get 0.499999999999997...
Solution 5:
this question was answered before on this site, you are doing an integer division, if you want to get the 0.5 use:
double x = (double)1/2;
and you will get the value of 0.5 .