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Order priority || Order of operations

I was working on some code in a class and came across the following:

int x 14; 
int y 3; 

x = x-- % y--'

Result after compilation 'x = 2' 'y = 2'

I am having a very difficult time understanding the order or operations for this particular scenario. My logic is based on the oracle operator precedence (here) http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html

Would conclude: x = (x = x -1)% (y = y - 1) (due to order priority)

Therefore: x = 13% 2

x = 1

y = 2

I'm wrong, tell me why. I have horses. Thanks in advance.

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3 answers


It:

int x = 2;
println(x--);

prints 2, but leaves x

at 1. Incrementing the suffix and decrementing gives you the value before changing the variable.

It:

int x = 2;
println(--x);

prints 1 and leaves x

at 1. Incrementing and decreasing the prefixes gives you the value after changing the variable.

EDIT:



If you assign x

in the same expression, the assignment occurs last.

int x = 3;
x = 2*(x--);

The value x--

is 3 (the value to is x

decremented). So after assignment x

ends up with a value of 6 in this case.

So for your example:

int x = 14;
int y = 3;

x = x-- % y--;

The value x--

is 14 (the value to is x

decremented). The value y--

is 3 (the value to is y

decremented). Therefore it is x

assigned 14%3==2

. y

remains with the reduced value, 2.

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x--

return x

and decrement after.

x = 14, y = 3
x = 14 % 3 → 2,  (x = x - 1 → 13 is done before the x receive  14 % 3 → 2)
y = y - 1 → 2


--x

reduce and return x

0


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I think this is happening:

int temp = (x % y);
x--;
y--;
x = temp;
0


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