Is “long x = 1/2” equal to 1 or 0, and why? [duplicate]

Answer 1

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...

Answer 2

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 .

Answer 3

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.

Answer 4

It's doing integer division, which truncates everything to the right of the decimal point.

Answer 5

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.

Answer 6

There are lots of different rounding conventions, the most common being rounding towards +inf, rounding towards -inf and rounding towards zero. Lots of people assume there's one right way, but they all have different ideas about what that one way should be ;-)

There is no intermediate non-integer result for integer division, but of course the division is done deterministically, and one particular rounding convention will always be followed for a particular platform and compiler.

With Visual C++ I get 5/2 = 2 and -5/2 = -2, rounding towards zero.

The rounding in C, C++ and Java is commonly called "truncation" - meaning drop off the unwanted bits. But this can be misleading. Using 4 bit 2s complement binary, doing what truncation implies gives...

 5/2 = 0101/0010 = 0010.1 --> 0010 =  2
-5/2 = 1011/0010 = 1101.1 --> 1101 = -3

Which is rounding towards -infinity, which is what Python does (or at least what it did in Python 2.5).

Truncation would be the right word if we used a sign-magnitude representation, but twos complement has been the de-facto standard for decades.

In C and C++, I expect while it's normally called truncation, in reality this detail is undefined in the standards and left to the implementation - an excuse for allowing the compiler to use the simplest and fastest method for the platform (what the processors division instruction naturally does). It's only an issue if you have negative numbers though - I've yet to see any language or implementation that would give 5/2 = 3.

I don't know what the Java standard says. The Python manual specifies "floor" division, which is a common term for rounding to -infinity.

EDIT

An extra note - by definition, if a/b = c remainder d, then a = (b*c)+d. For this to hold, you have to choose a remainder to suite your rounding convention.

People tend to assume that remainders and modulos are the same, but WRT signed values, they can be different - depending on the rounding rules. Modulo values are by definition never negative, but remainders can be negative.

I suspect the Python round-towards-negative-infinity rule is intended to ensure that the single % operator is valid both as a remainder and as a modulo. In C and C++, what % means (remainder or modulo) is (yes, you guessed it) implementation defined.

Ada actually has two separate operators - mod and rem. With division required to round towards zero, so that mod and rem do give different results.