Unreachable code while loop

Question

When I'm compiling this code

public static void main(String [] args) {

        int x = 0;        

        while(false)
        {                        
            System.out.println(hello);
        }
    }

it is showing compile time error unreachable code.

But when I modified this code to

public static void main(String [] args) {

        int x = 0;        
        boolean result = false;
        while(result)
        {                        
            x=4;
        }
    }

it's working fine.

Can somebody tell me the reason behind this behavior.

Whilst this is Java, see also: https://communities.coverity.com/blogs/development-testing-blog/2013/11/06/c-reachability - I suspect the reasoning is similar. — Rowland Shaw, Jan 04 '14 at 15:31

score 9 · Answer 1 · edited May 23 '17 at 11:59

It is because boolean result = false is not a constant expression whereas false is. If you try the code below, it won't compile either because result is now a constant:

final boolean result = false;
while(result) { x=4; }

However this would compile, because result is not a constant variable any longer:

final boolean result;
result = false;
while(result) { x=4; }

See also: Why does the Java compiler not understand this variable is always initialized? for a similar discussion.

Juned Ahsan · Answer 2 · 2014-01-04T15:07:41.227

6

Usage of constant false in the following statement

  while(false)

is resolved to false at compilation time hence compiler complains for unreachable code.

But when you use a variable instead:

    boolean result = false;
    while(result)

compiler is not sure about the value of it at compile time and hence does not complain.

edited Jan 04 '14 at 15:07

answered Jan 04 '14 at 14:47

Juned Ahsan

67,789
12
98
136

Maybe too advanced for the asker, but it is possible to memory-modify the value of `result` to _true_ just before `while(result)` is called, resulting a _hacked_ execution where the code inside the while loop is reached. This cannot happen in `while(false)`, hence the code is truly unreachable (compiler ignores it, doesn't even convert it into bytecode). – ADTC Jan 04 '14 at 14:55

score 4 · Answer 3 · answered Jan 04 '14 at 14:50

The behavior of the compiler is precisely specified in the Java Language Specification, Section 14.21. Unreachable Statements.

Here is a key quote, which directly addresses your question:

This section is devoted to a precise explanation of the word "reachable." The idea is that there must be some possible execution path from the beginning of the constructor, method, instance initializer, or static initializer that contains the statement to the statement itself. The analysis takes into account the structure of statements. Except for the special treatment of while, do, and for statements whose condition expression has the constant value true, the values of expressions are not taken into account in the flow analysis.

score 1 · Answer 4 · answered Jan 04 '14 at 14:51

1

Java uses a simple flow analysis algorithm to find most common cases of unreachable code, and all such unreachable code blocks will be flagged as compile-time errors. That's why your "while (false) { ... }" statement produces an error.

answered Jan 04 '14 at 14:51

Ashish

1,121
2
15
25

Unreachable code while loop

4 Answers4

Linked