Assignment Operators Problems Solutions

x++;

x++ is an example of post-increment. If x++ appears in an expression, we would use the current value of x in the expression, and then increment x afterwards. The same would be true in the case of post-decrement. In this case, however, there is no expression, so we simply increment x from 10 to 11. Used this way, x++; and ++x; are equivalent.

--y:

--y; is an example of pre-decrement. If --y appears in an expression, we would first decrement y and use the new value of y in the expression. The same would be true in the case of pre-increment. In this case,however, there is no expression to evaluate, so we simply decrement y from 15 to 14. Used this way (when there is no expression), --y; and y--; are equivalent.

a = ++z - 5;

First, look at the expression on the right for any pre-increment and/or pre-decrement operators and perform them first.
In this case, ++z is a pre-increment, so we increment z from 6 to 7.
We then evaluate the expression... 7 - 5 is 2, which gets stored in a.

b = 16 - y++;

Since the expression on the right has no pre-increment or pre-decrement operators,
we use the current value of y to evaluate the expression.
16 - 15 = 1. So 1 is stored in b, then y is incremented from 15 to 16.

c = ++x + z--;

Since ++x is pre-increment, we first increment x from 10 to 11 and use that value in the expression.
Since z-- is post-decrement, we use the current value of z (which is 6) in the expression,
then decrement z from 6 to 5. So the steps are:

1. increment x from 10 to 11
2. evaluate 11 + 6, which is 17 and store in c
3. decrement z from 6 to 5

x += 2 * y;

x += 2 * y; is equivalent to x = x + (2 * y);
Since y = 15, x = 10 + (2 * 15) = 10 + 30 = 40

y -= x / --z;

We need to evaluate x / --z, then subtract that answer from y to get the new value of y.
We know that "--" is the decrement operator, but we must pay attention to whether it's used as pre-decrement or post-decrement.
In this case --z is pre-decrement, so we decrement z from 6 to 5 first,
then we use the new value of z and evaluate x / 5.
Since x is 10, x / 5 = 10 / 5 = 2.
Finally, y -= 2; is equivalent to y = y - 2; so y = 15 - 2 which is 13.

z += x-- + 5;

We first need to evaluate x-- + 5, then add that value to z
The fundamental question is whether we decrement x before (pre-decrement) or after (post-decrement) we use x's value in the calculation.
In this case, x-- is post-decrement, so we calculate x-- + 5 as 10 + 5 which is 15
then we decrement x from 10 to 9.
Finally, z += 15 is equivalent to z = z + 15.
Since z is currently 6, we get z = 6 + 15 which is 21.

y /= z + 2;

This is one of the simpler problems, but reminds us about integer division.
Since z is 6, z + 2 = 8.
y /= 8 is equivalent to y = y / 8.
Since the value of y is currently 15, this becomes y = 15 / 8, which is 1
because we are doing integer division.

x *= ++y - z--;

In order to evaluate ++y - z--, we must decide which values of y and z to use.....
do we use the current values, or do we increment (decrement) first and use the new values.
In this problem it's both. Because ++y is pre-increment, we first increment y from 15 to 16.
Next look at z--. This is post-increment, so we use the current value of z, which is 6, then decrement z.
So the steps are:

1. increment y from 15 to 16
2. use the current value of z (which is 6) and calculate 16 - 6 = 10 as the value of the expression
3. calculate x *= 10 which is equivalent to x = x * 10. Since x is currently 10, this evaluates to x = 10 * 10 which is 100
4. now decrement z from 6 to 5