The following code in Java:
int a = 0, b = 0, c = 0;
boolean d = (a++ > 0 && b-- < 0) || --c < 0;
results in the values:
a = 1, b = 0, c = -1 and d = true
I don't understand why a is = 1, because it is a post-increment and should also react the same way that value b does. Also, if I change the b-- to --b it still has no effect on the value of b.
What is the best way of understanding this logic?
a++ > 0 returns false, since a++ return the previous value of a (0).
Therefore b-- < 0 is not evaluated at all, since && is a short circuiting operator. The right operand is only evaluated if the left operand is true.
--c < 0 is evaluated, since the first operand of the || operator is false, so the second operand must be evaluated.
After d is evaluated, the value of a is 1, since a was incremented. b remains 0, since b-- wasn't executed. c is -1 since --c was executed.
And d is true since --c < 0 is true.
a++ at moment of comparing is NOT >0, so b part is not evaluated, and not incremented.
Second part of && and || expressions is not evaluated if cannot change result. For && first false determines result false.
Analogically, first true determines result of || (but not here)
After the expression runs then a will indeed be 1 as the post increment operator has then executed.
During the evaluation of the expression a is 0 as far as the expression is concerned (but a is now 1 for other code), so the first part of the logical OR, your logical AND, is evaluated as false and the b post decrement does not get evaluated.
The second part of the logical or then runs, and the pre-decrement executes to reduce c to -1, which then causes the comparison to evaluate to true, hence d being true.
Expand the different clauses to separate variables and run through a debugger for a more interactive explanation of what's happening.
When you write a++ OR b--, it passes the value 'a' and 'b' currently holds, and then modifies the variables.
So your expression is simply,
boolean d = (0 > 0 && 0 < 0) || -1 < 0;
Now while evaluating, (0 > 0 && 0 < 0) -> first expression returns false and you have used short-circuit AND operator. Meaning don't evaluate the right hand side if it isn't necessary. As LHS of && returns false, b-- won't be evaluated.
And we have 'd' is true. Remember,
Post-increment within expressions --> assign current value first, modify later.
Pre-increment within expressions --> modify first, assign updated value later.
Refer this link for better understanding short circuit operators: Short-Circuit Explanation
a++ effectively means a= a+1 So, the existing value of a (=0) is used for evaluating the expression, and later on the value of a is incremented. So, a becomes 1.
Regarding why b =0 and why --b has the same effect as b--, the value of b is unchanged because of &&. Since a++ > 0 is not satisfied, the next expression b-- < 0 is not evaluated and value of b remains 0. && stops evaluation once the expression evaluates to false, while || stops evaluation once the statement evaluates to false.
Please Execute This code..
int a = 0, b = 0, c = 0;
System.out.println("b = " + --b);
b=0;
System.out.println("b = " + b--);
System.out.println("a = " + a++);
a=0;
System.out.println("a = " + ++a);
System.out.println("c = " + c--);
c=0;
System.out.println("c = " + --c);
a=b=c=0;
boolean d = (a++ > 0 && --b < 0) || --c < 0;
System.out.println("d = " + d);
I think you understand full logic.
