AND/OR (&&/||) logic for multiple condition statements [closed]

Question

If you have an if-statement in C# that checks multiple conditions:

if (a == 5 && b == 9) { ... }

Does b == 9 still get checked if a == 5 condition is false, or does it automatically exit since there's no way this could pass anymore?

Similarly, for an OR if-statement:

if (a == 5 || b == 9) { ... }

Will b == 9 still get checked if a == 5 is true?

Answer 1

Both && and || is "short-circuiting" operators, which means that if the answer is known from the left operand, the right operand is not evaluated.

This means that:

a && b

b will not be evaluated if a is false, since the final answer is already known.

Likewise:

a || b

b will not be evaluated if a is true, since the final answer is already known.

If you want both operands to be evaluated, use the & and | operators instead.

The bonus of this is that you can write expressions that would fail if all operands was evaluated. Here's a typical if-statement:

if (a != null && a.SomeProperty != null && a.SomeProperty.Inner != null)
    ... use a.SomeProperty.Inner

If a was null, and the expression would go on to evaluate a.SomeProperty, it would throw a NullReferenceException, but since && short-circuits, if a is null, the expression will not evaluate the rest and thus not throw the exception.

Obviously, if you replace && with &, it will throw that exception if either a or a.SomeProperty is null.

Answer 2

Conceptually, && and || short-circuit.

But since you don't have any side-effects there, the JIT compiler is free to remove the short-circuiting. I don't know whether it actually does so or not.

Answer 3

For : if (a == 5 && b == 9) { ... }

Does b == 9 still get checked if a == 5 condition is false, or does it automatically exit since there's no way this could pass anymore?

If a == 5 is false no any other control will be executed on that line.

For: if (a == 5 || b == 9) { ... }

Will b == 9 still get checked if a == 5 is true?

Pass inside immediately, as first condition already satisfies requirements.