What is the result stored in x, after evaluating the following code snippet?int x =-5;X= X++ %2+3 *--x +2:
Question
What is the result stored in x, after evaluating the following code snippet?
int x = -5;
X = X++ % 2 + 3 * --x + 2;
Solution
The code snippet you provided seems to be written in Java or a similar language. Let's break it down step by step:
-
int x = -5;
This line initializes an integer variablex
with a value of-5
. -
x = x++ % 2 + 3 * --x + 2;
This line is a bit more complex, so let's break it down further:-
x++
This is a post-increment operation. It means thatx
is used in its current state (which is-5
), and then incremented by1
after the operation. So, for this operation,x
is-5
, but immediately after it becomes-4
. -
--x
This is a pre-decrement operation. It means thatx
is decremented by1
before it's used in the operation. Sincex
was-4
after the post-increment, it now becomes-5
again before this operation. -
%
This is the modulus operator. It returns the remainder of the division of the number to the left by the number on the right.-5 % 2
equals-1
because-5
divided by2
equals-2
with a remainder of-1
. -
*
This is the multiplication operator.3 * -5
equals-15
. -
+
This is the addition operator.-1 + -15 + 2
equals-14
.
-
So, after evaluating the entire code snippet, x
should be -14
.
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