问题
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?
回答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)
回答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.
回答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 &.
回答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
来源：https://stackoverflow.com/questions/2028653/boolean-expression-order-of-evaluation-in-java