Boolean expression order of evaluation in Java?

Question

Suppose I have the following expression

String myString = getStringFromSomeExternalSource();
if (myString != null && myString.trim().length() != 0) {
...
}

Eclipse warns me that myString might be null in the second phrase of the boolean expression. However, I know some that some compilers will exit the boolean expression entirely if the first condition fails. Is this true with Java? Or is the order of evaluation not guaranteed?

Solution 1

However, I know some that some compilers will exit the boolean expression entirely if the first condition fails. Is this true with Java?

Yes, that is known as Short-Circuit evaluation.Operators like && and || are operators that perform such operations.

Or is the order of evaluation not guaranteed?

No,the order of evaluation is guaranteed(from left to right)

Solution 2

Java should be evaluating your statements from left to right. It uses a mechanism known as short-circuit evaluation to prevent the second, third, and nth conditions from being tested if the first is false.

So, if your expression is myContainer != null && myContainer.Contains(myObject) and myContainer is null, the second condition, myContainer.Contains(myObject) will not be evaluated.

Edit: As someone else mentioned, Java in particular does have both short-circuit and non-short-circuit operators for boolean conditions. Using && will trigger short-circuit evaluation, and & will not.

Solution 3

James and Ed are correct. If you come across a case in which you would like all expressions to be evaluated regardless of previous failed conditions, you can use the non-short-circuiting boolean operator &.

Solution 4

Yes, Java practices lazy evaluation of if statements in this way. if myString==null, the rest of the if statement will not be evaluated

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