I was doing a practice Computer Science UIL test form when I came across this problem:
What is output by the following?
int a = 5;
int b = 7;
int c = 10;
c = b+++-c--+--a;
System.out.println(a + " " + b + " " + c);
I put down the answer "No output due to syntax error" but I got it wrong. The real answer was 4 8 1! (I tested it out myself)
Can someone please explain to me how line 4 works?
Thanks
I added some parentheses:
int a = 5;
int b = 7;
int c = 10;
c = (b++) + (-(c--)) + (--a);
System.out.println(a + " " + b + " " + c);
b++ : b = b + 1 after b is used
c-- : c = c - 1 after c is used
--a : a = a - 1 before a is used
Look at it like this:
(b++) + (-(c--)) + (--a)
That should make more sense!
Look at Operator Precedence to see why it works this way.
Break up the statement a bit. It's intentionally obfuscated.
c = b++ + -c-- + --a;
What this means:
c is assigned the result of...
b (incrementation will take effect after this line), plus- of c (decrementation will take effect after this line), plusa (decrementation takes immediate effect).Replace the variables with the values, and you get:
c = 7 + (-10) + 4
c = 1
...and the result of your print statement should be:
4 8 0
Let's slow down, and look hard at the equation. Think about this carefully.
int a = 5;
int b = 7;
int c = 10;
c = b+++-c--+--a;
b++ means increment b after assignment, so b stays equal to its original value in the equation, but will be incremented afterward the equation.
Then there's a +.
Then a negated c--. c gets decremented but will remain the same for the equation.
Then add that to --a, which means a gets decremented immediately.
So the variables values at the print statement will be:
c = 7 + -10 + 4 = 1
a = 4
b = 8
May I add that in my opinion this is a bad question for a test. All its really asking is if you understand i++ vs ++i.
