if x = 3 and z is unassigned,
why does z = x-- - --x evaluates to 2?

my professor is lecturing about this at the moment, and I'm currently stuck with this dilemma. Unfortunately, no one can explain why it happens.

if x = 3 and z is unassigned,
why does z = x-- - --x evaluates to 2?

my professor is lecturing about this at the moment, and I'm currently stuck with this dilemma. Unfortunately, no one can explain why it happens.

• c#
• javascript

• x in post decrement minus x pre decrement – arscariosus Commented Jun 14, 2012 at 7:49
• 4 is this C# or this is a JavaScript? Choose correct tag, please. – Tigran Commented Jun 14, 2012 at 7:49
• 1 Duane: The font makes it look weird. Its x-- (x minus minus) minus --x (minus minus x) – Matt Commented Jun 14, 2012 at 7:49
• 2 very bad practice to have such code lines. There are plenty of other ways to explain the usage of x-- and --x differences. – Steve B Commented Jun 14, 2012 at 7:55
• 2 Nobody would ever write something like this, I wouldn't use it to explain pre/post decrement, but I'd certainly use it to see if the students really understood the concept – Francesco Baruchelli Commented Jun 14, 2012 at 8:06

3 Answers 3

on x--, x = 3, and after that it's 2. on --x, x = 1, because substraction (from 2) is done beforehand.

Therefore, 3 - 1 = 2.

Here is the order of operations, illustrated for better understanding:

• x-- - --x Hold value of x (lets call it tmpA). tmpA is 3.
• x-- - --x Decreases x. It is now 2.
• x-- - --x Decreases x. It is now 1.
• x-- - --x Hold value of x (lets call it tmpB). tmpB is 1.
• x-- - --x Performs substruction from calculated values. 3 - 1 = 2.

The -- prefix means the decrement will be done before evaluating the expression and the postfix -- means the decrement will be done after evaluation the expression.

Ok, its pretty simple:

let's add brackets:

z = ( x-- ) - ( --x )

^^ this is how compiler sees your code after tokenizing.

Compiler evaluates equation (right part) from left to right

Now,

x-- is equal to POP the value of x and then decrement it and PUSH back value into a memory. Ok, lets do it:

Current value of X is 3, decremented is 2 - so, in equation we'll get 3, but X will contain new value 2.

--x is equal to decrement X value and then POP this value into equation. Let's do it:

Current value of X is 2 (because previous operation decremented it), and now we want to decrease it once again. 2-1 = 1, got it.

Now, back to whole equation: z = (3) - (1) = 2.