Confusing while block

Question

The following code block gives me a compile time error.

        while(false)
        {
          System.out.println("HI");
        }

The error says that there is an unreachable statement. BUT the following code compiles

   boolean b=false;
   while(b)
   {
        System.out.println("Hi");
    }

All i could think of was this -> In case-1 as false is a literal so the compiler finds that its unreachable and in case 2 variable b in while condition block is checked at runtime so there is no compilation error?

Patricia Shanahan · Answer 1 · 2013-09-12T00:04:45.040

The compiler writers do not get to decide which conditions to flag as errors - the rules are in the Java Language Specification, for this issue in 14.21 Unreachable Statements.

The relevant sentence is: "The contained statement is reachable iff the while statement is reachable and the condition expression is not a constant expression whose value is false."

In each case, you have a reachable while statement. In the first version, false is a constant expression whose value is false, so the contained statement is not considered reachable. In the second version, b is not a constant expression at all, so the contained statement is treated as being reachable.

Adding final to the declaration of b changes the while condition to a constant expression whose value is false, making the contained statement unreachable again.

Specifying the rules for what is and is not a compile time error in the JLS has the benefit that all Java compilers should accept the same set of programs. The rules generally do not require the compiler to do data flow analysis, presumably to limit the cost and difficulty of writing compilers.

Jordan · Accepted Answer · 2013-09-11T23:41:03.053

The compiler sees while(false), which will never be true, so you cannot reach the println. This throws an error.

Meanwhile, although while(b) will never be true either, the compiler doesn't automatically know this, because b isn't automatically false, it is a boolean that happens to have the value false, it could change at some point. (It doesn't here, but it could have).

To make this point more general, the compiler will look at the type of a variable, but not what the variable actually is. Many beginning programming classes have sections which deal with how polymorphism and type casting leads to run-time errors in some cases, and compiler errors in others. If you happen to have taken a course such as these, you can think of your question here as having a similar explanation.

score 2 · Answer 3 · answered Sep 11 '13 at 23:35

2

This makes sense to me. While (false) will never evaluate successfully, whereas in the case of while(b) - the value of b could be updated to true at some point in the program's lifetime.

answered Sep 11 '13 at 23:35

Davie Brown

717
4
23

score 1 · Answer 4 · answered Sep 11 '13 at 23:32

The compiler isn't extremely smart about things. It'll find a block where you have a provably false condition, and it'll cry. Give it something like this, and javac isn;t wired to look even that far. Tings obvious to humans are not necessarily obvious to computers, and vice versa.

score 1 · Answer 5 · answered Sep 24 '13 at 08:50

1

Compiler sees while(false) and can determine that block inside while won't execute ever.

while in case of boolean, value of b evaluated at runtime so it won't stop you

answered Sep 24 '13 at 08:50

Dax Joshi

143
11

Confusing while block

5 Answers5