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Unreachable statement compile error in Java [duplicate]

This question already has answers here:
Unreachable code: error or warning? [closed]
(12 answers)
Closed 9 years ago.
class For1
{
public static void main(String args[])
{
int a = 0;
for(;;)
{
break;
System.out.println(a); //Line 1
++a;//Line 2
}
}
}
I know that Line 1/Line 2 will never be executed.
But still I don't understand why a compile time error is thrown.
I am getting "unreachable statement" compile error.
Does it mean that compiler checks whether it is able to compile for all branches/lines of code ?

Does it mean that compiler checks whether it is able to compile for all branches/lines of code ?
It means the compiler checks that every statement is reachable.
From section 14.21 of the JLS:
It is a compile-time error if a statement cannot be executed because it is unreachable.
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.
The section then documents how reachability is defined.
In particular, the relevant points in your case are:
Every other statement S in a non-empty block that is not a switch block is reachable iff the statement preceding S can complete normally.
A break, continue, return, or throw statement cannot complete normally.
So your "line 1" statement is preceded by a statement (break;) which cannot complete normally, and therefore it's unreachable.

The compiler is also able to make that conclusion, and assumes you are making a mistake. And yes, the Java compiler does a pretty good amount of "Data-Flow Analysis". The most common related message is the one about variables not initialized. The second most frequent is, I believe, precisely this one, about code not reachable.

Does it mean that compiler checks whether it is able to compile for
all branches/lines of code ?
Yes compiler compiles the whole body of code and make byte code according to your code, it smarter enough to detects unreachable code also dead code. Immediate break in the for-loop makes unreachable other statements.
for(;;){
break;
... // unreachable statement
}
int i=1;
if(i==1)
...
else
... // dead code

Unreachable code is meaningless and redundant. If you have some unreachable code in your program it is a mistake and needs to be fixed. Hence compiler throws an error.
You can refer to similar questions below
Unreachable code: error or warning?
and
Why does Java have an "unreachable statement" compiler error?

The compiler is able to determine that these two statement will never, ever be executed, and helps you write correct code by refusing to compile it, because this has 99.9% chance of being an error rather than a conscious choice to add statements that will never be executed.

The compiler will check if there is more code after certain keywords. Another keyword which will cause a similar message is if you replace break by return.

Missing return statement in a non-void method compiles

I encountered a situation where a non-void method is missing a return statement and the code still compiles.
I know that the statements after the while loop are unreachable (dead code) and would never be executed. But why doesn't the compiler even warn about returning something? Or why would a language allow us to have a non-void method having an infinite loop and not returning anything?
public int doNotReturnAnything() {
while(true) {
//do something
}
//no return statement
}
If I add a break statement (even a conditional one) in the while loop, the compiler complains of the infamous errors: Method does not return a value in Eclipse and Not all code paths return a value in Visual Studio.
public int doNotReturnAnything() {
while(true) {
if(mustReturn) break;
//do something
}
//no return statement
}
This is true of both Java and C#.

Why would a language allow us to have a non-void method having an infinite loop and not returning anything?
The rule for non-void methods is every code path that returns must return a value, and that rule is satisfied in your program: zero out of zero code paths that return do return a value. The rule is not "every non-void method must have a code path that returns".
This enables you to write stub-methods like:
IEnumerator IEnumerable.GetEnumerator()
{
throw new NotImplementedException();
}
That's a non-void method. It has to be a non-void method in order to satisfy the interface. But it seems silly to make this implementation illegal because it does not return anything.
That your method has an unreachable end point because of a goto (remember, a while(true) is just a more pleasant way to write goto) instead of a throw (which is another form of goto) is not relevant.
Why doesn't the compiler even warn about returning something?
Because the compiler has no good evidence that the code is wrong. Someone wrote while(true) and it seems likely that the person who did that knew what they were doing.
Where can I read more about reachability analysis in C#?
See my articles on the subject, here:
ATBG: de facto and de jure reachability
And you might also consider reading the C# specification.

The Java compiler is smart enough to find the unreachable code ( the code after while loop)
and since its unreachable, there is no point in adding a return statement there (after while ends)
same goes with conditional if
public int get() {
if(someBoolean) {
return 10;
}
else {
return 5;
}
// there is no need of say, return 11 here;
}
since the boolean condition someBoolean can only evaluate to either true or false, there is no need to provide a return explicitly after if-else, because that code is unreachable, and Java does not complain about it.

The compiler knows that the while loop will never stop executing, hence the method will never finish, hence a return statement is not necessary.

Given your loop is executing on a constant - the compiler knows that it's an infinite loop - meaning the method could never return, anyway.
If you use a variable - the compiler will enforce the rule:
This won't compile:
// Define other methods and classes here
public int doNotReturnAnything() {
var x = true;
while(x == true) {
//do something
}
//no return statement - won't compile
}

The Java specification defines a concept called Unreachable statements. You are not allowed to have an unreachable statement in your code (it's a compile time error). You are not even allowed to have a return statement after the while(true); statement in Java. A while(true); statement makes the following statements unreachable by definition, therefore you don't need a return statement.
Note that while Halting problem is undecidable in generic case, the definition of Unreachable Statement is more strict than just halting. It's deciding very specific cases where a program definitely does not halt. The compiler is theoretically not able to detect all infinite loops and unreachable statements but it has to detect specific cases defined in the specification (for example, the while(true) case)

The compiler is smart enough to find out that your while loop is infinite.
So the compiler cannot think for you. It cannot guess why you wrote that code. Same stands for the return values of methods. Java won't complain if you don't do anything with method's return values.
So, to answer your question:
The compiler analyzes your code and after finding out that no execution path leads to falling off the end of the function it finishes with OK.
There may be legitimate reasons for an infinite loop. For example a lot of apps use an infinite main loop. Another example is a web server which may indefinitely wait for requests.

In type theory, there is something called the bottom type which is a subclass of every other type (!) and is used to indicate non-termination among other things. (Exceptions can count as a type of non-termination--you don't terminate via the normal path.)
So from a theoretical perspective, these statements that are non-terminating can be considered to return something of Bottom type, which is a subtype of int, so you do (kind of) get your return value after all from a type perspective. And it's perfectly okay that it doesn't make any sense that one type can be a subclass of everything else including int because you never actually return one.
In any case, via explicit type theory or not, compilers (compiler writers) recognize that asking for a return value after a non-terminating statement is silly: there is no possible case in which you could need that value. (It can be nice to have your compiler warn you when it knows something won't terminate but it looks like you want it to return something. But that's better left for style-checkers a la lint, since maybe you need the type signature that way for some other reason (e.g. subclassing) but you really want non-termination.)

There is no situation in which the function can reach its end without returning an appropriate value. Therefore, there is nothing for the compiler to complain about.

Visual studio has the smart engine to detect if you have typed a return type then it should have a return statement with in the function/method.
As in PHP Your return type is true if you have not returned anything. compiler get 1 if nothing has returned.
As of this
public int doNotReturnAnything() {
while(true) {
//do something
}
//no return statement
}
Compiler know that while statement itself has a infinte nature so not to consider it. and php compiler will automatically get true if you write a condition in expression of while.
But not in the case of VS it will return you a error in the stack .

Your while loop will run forever and hence won't come outside while; it will continue to execute. Hence, the outside part of while{} is unreachable and there is not point in writing return or not. The compiler is intelligent enough to figure out what part is reachable and what part isn't.
Example:
public int xyz(){
boolean x=true;
while(x==true){
// do something
}
// no return statement
}
The above code won't compile, because there can be a case that the value of variable x is modified inside the body of while loop. So this makes the outside part of while loop reachable! And hence compiler will throw an error 'no return statement found'.
The compiler is not intelligent enough (or rather lazy ;) ) to figure out that whether the value of x will be modified or not. Hope this clears everything.

"Why doesn't the compiler even warn about returning something? Or why would a language allow us to have a non-void method having an infinite loop and not returning anything?".
This code is valid in all other languages too (probably except Haskell!). Because the first assumption is we are "intentionally" writing some code.
And there are situations that this code can be totally valid like if you are going to use it as a thread; or if it was returning a Task<int>, you could do some error checking based on the returned int value - which should not be returned.

I may be wrong but some debuggers allow modification of variables. Here while x is not modified by code and it will be optimized out by JIT one might modify x to false and method should return something (if such thing is allowed by C# debugger).

The specifics of the Java case for this (which are probably very similar to the C# case) are to do with how the Java compiler determines if a method is able to return.
Specifically, the rules are that a method with a return type must not be able to complete normally and must instead always complete abruptly (abruptly here indicating via a return statement or an exception) per JLS 8.4.7.
If a method is declared to have a return type, then a compile-time
error occurs if the body of the method can complete normally.
In other words, a method with a return type must return only by using
a return statement that provides a value return; it is not allowed to
"drop off the end of its body".
The compiler looks to see whether normal termination is possible based on the rules defined in JLS 14.21 Unreachable Statements as it also defines the rules for normal completion.
Notably, the rules for unreachable statements make a special case just for loops that have a defined true constant expression:
A while statement can complete normally iff at least one of the
following is true:
The while statement is reachable and the condition expression is not a
constant expression (§15.28) with value true.
There is a reachable break statement that exits the while statement.
So if the while statement can complete normally, then a return statement below it is necessary since the code is deemed reachable, and any while loop without a reachable break statement or constant true expression is considered able to complete normally.
These rules mean that your while statement with a constant true expression and without a break is never considered to complete normally, and so any code below it is never considered to be reachable. The end of the method is below the loop, and since everything below the loop is unreachable, so is the end of the method, and thus the method cannot possibly complete normally (which is what the complier looks for).
if statements, on the other hand, do not have the special exemption regarding constant expressions that are afforded to loops.
Compare:
// I have a compiler error!
public boolean testReturn()
{
final boolean condition = true;
if (condition) return true;
}
With:
// I compile just fine!
public boolean testReturn()
{
final boolean condition = true;
while (condition)
{
return true;
}
}
The reason for the distinction is quite interesting, and is due to the desire to allow for conditional compilation flags that do not cause compiler errors (from the JLS):
One might expect the if statement to be handled in the following
manner:
An if-then statement can complete normally iff at least one of the
following is true:
The if-then statement is reachable and the condition expression is not
a constant expression whose value is true.
The then-statement can complete normally.
The then-statement is reachable iff the if-then statement is reachable
and the condition expression is not a constant expression whose value
is false.
An if-then-else statement can complete normally iff the then-statement
can complete normally or the else-statement can complete normally.
The then-statement is reachable iff the if-then-else statement is
reachable and the condition expression is not a constant expression
whose value is false.
The else-statement is reachable iff the if-then-else statement is
reachable and the condition expression is not a constant expression
whose value is true.
This approach would be consistent with the treatment of other control
structures. However, in order to allow the if statement to be used
conveniently for "conditional compilation" purposes, the actual rules
differ.
As an example, the following statement results in a compile-time
error:
while (false) { x=3; } because the statement x=3; is not reachable;
but the superficially similar case:
if (false) { x=3; } does not result in a compile-time error. An
optimizing compiler may realize that the statement x=3; will never be
executed and may choose to omit the code for that statement from the
generated class file, but the statement x=3; is not regarded as
"unreachable" in the technical sense specified here.
The rationale for this differing treatment is to allow programmers to
define "flag variables" such as:
static final boolean DEBUG = false; and then write code such as:
if (DEBUG) { x=3; } The idea is that it should be possible to change
the value of DEBUG from false to true or from true to false and then
compile the code correctly with no other changes to the program text.
Why does the conditional break statement result in a compiler error?
As quoted in the loop reachability rules, a while loop can also complete normally if it contains a reachable break statement. Since the rules for the reachability of an if statement's then clause do not take the condition of the if into consideration at all, such a conditional if statement's then clause is always considered reachable.
If the break is reachable, then the code after the loop is once again also considered reachable. Since there is no reachable code that results in abrupt termination after the loop, the method is then considered able to complete normally, and so the compiler flags it as an error.

Why is return needed even after System.exit(0);

Consider this function :
public boolean foo(){
System.exit(1);
//The lines beyond this will not be read
int bar = 1; //L1
//But the return statement is required for syntactically correct code
return false; //L2
//error here for unreachable code
//int unreachable = 3; //L3
}
Can someone please explain why L1 and L2 visibly not reachable does not give warnings but L3 does.

Because as far as the compiler is concerned, System.exit() is just another method call.
The fact that what it does is end the process can only be found out from the implementation (which is native code, not that it makes any difference).
If you have to put System.exit() in your code (usually it's best to avoid it, unless you want to return a code other than 0), it should really be in a method that returns void, main() for example. It's nicer that way.
As for the reachability, the explanation is the same: return is a keyword of the Java language, so the compiler or the parser the IDE uses can tell that it's theoretically impossible for code after the return statement to be executed. These rules are defined here.

The Java compiler doesn't know anything about System.exit. It's just a method as far as it's concerned - so the end of the statement is reachable.
You say that L1 and L2 are "visibly not reachable" but that's only because you know what System.exit does. The language doesn't - whereas it does know what a return statement does, so it knows that L3 really isn't reachable.
I sometimes think it would be useful to be able to declare that a method isn't just void, but never terminates normally - it never just returns (although it may throw an exception). The compiler would then be able to use that information to make the end of any calling expression unreachable, preventing this sort of thing from being a problem. However, that's just my dreams around language design - Java doesn't have anything similar, and it would be a very bad idea for the compiler to "know" that particular JRE methods will never return normally, when that concept can't be expressed directly within the language.
Instead, the compiler is bound by the rules of section 14.21 of the JLS, including:
The first statement in a non-empty block that is not a switch block is reachable iff the block is reachable.
Every other statement S in a non-empty block that is not a switch block is reachable iff the statement preceding S can complete normally.
...
An expression statement can complete normally iff it is reachable.
(A method call is an expression statement.)
Then from section 8.4.7:
If a method is declared to have a return type, then a compile-time error occurs if the body of the method can complete normally (§14.1).
and in 14.1:
Unless otherwise specified, a statement completes normally if all expressions it evaluates and all substatements it executes complete normally.
So the call to System.exit() can complete normally as far as the compiler is concerned, which means the body of the foo method can complete normally, which leads to the error.

Me: "Can anything be executed after a return statement?"
Java: "NO."
Me: "Can anything be executed after I call System.exit?"
Java: "That's a method, I don't know what it does - it's not a reserved keyword, it doesn't affect program flow as far as I know" (and it may not even work (I don't know if exit can throw exceptions (or future variants of it)))

From the language standpoint, there are only 2 ways to escape current scope: return and throw. Method calls are never considered the same way, even if they consist of the only line of code:
void method() {
throw new RuntimeException();
}
Even more. In theory, any method call can cause RuntimeException to be thrown. In this case, compiler should probably give you warnings for absolutely any method call which is not inside a try block:
...
method(); // WARNING: may throw in theory
anotherMethod(); // WARNING: possible unreachable code, also may throw
anotherMethod2(); // WARNING: possible unreachable code, also may throw
// etc
...
For your question logic is the same.

If a method is declared to return a non-void value, then it must contain a return statement somewhere, even if it's never reached (like in the code in the question).
From the compiler's point of view, System.exit() is just another method call, with nothing special about it that indicates that the program ends as soon as it's reached. Only you, as the programmer, know this fact - but it's something outside of the compiler's knowledge.
About the second part of your question - nothing can go after a return statement in a block of code inside a method, as that will always be unreacheable code. That explains why the compiler complains about the L3 line.

I know this does not make sense, and still this is how the Java compiler works, when you're calling a method.
The compiler does not know at this point what Sytem.exist does (why should rt.jar be different than other jars you compile with , in that sense?).
This is in contrast for example to the next piece of code -
public int test() {
throw new NullPointerException("aaaa");
}
Where the compiler can tell that an exception is always thrown, and therefore no return is needed.

The compiler checks whether some code is reachable only with respect to the return keyword (also throw and break(in case of loops), in general). For the compiler the exit method call is just another call, it doesn't know its meaning, so it doesn't know that the code afterwards will never be reached.

It's possible that the static code analysis tool is not taking into account that the application is terminating.

Unreachable code compiler error [duplicate]

This question already has answers here:
Why does Java have an "unreachable statement" compiler error?
(8 answers)
Closed 6 years ago.
The following code gives an unreachable statement compiler error
public static void main(String[] args) {
return;
System.out.println("unreachable");
}
Sometimes for testing purposes a want to prevent a method from being called, so a quick way to do it (instead of commenting it out everywhere it's used) is to return immediately from the method so that the method does nothing. What I then always do to get arround the compiler error is this
public static void main(String[] args) {
if (true) {
return;
}
System.out.println("unreachable");
}
I'm just curious, why is it a compiler error?? Will it break the Java bytecode somehow, is it to protect the programmer or is it something else?
Also (and this to me is more interesting), if compiling java to bytecode does any kind of optimization (or even if it doesn't) then why won't it detect the blatant unreachable code in the second example? What would the compiler pseudo code be for checking if a statement is unreachable?

Unreachable code is meaningless, so the compile-time error is helpful. The reason why it won’t be detected at the second example is, like you expect, for testing / debugging purposes. It’s explained in The Specification:
if (false) { x=3; }
does not result in a compile-time error. An optimizing compiler may
realize that the statement x=3; will never be executed and may choose
to omit the code for that statement from the generated class file, but
the statement x=3; is not regarded as "unreachable" in the technical
sense specified here.
The rationale for this differing treatment is to allow programmers to
define "flag variables" such as:
static final boolean DEBUG = false;
and then write code such as:
if (DEBUG) { x=3; }
The idea is that it should be possible to change the value of DEBUG
from false to true or from true to false and then compile the code
correctly with no other changes to the program text.
Reference: http://docs.oracle.com/javase/specs/jls/se8/html/jls-14.html#jls-14.21

Its because the compiler writer assumed that the human at the controls is dumb, and probably didn't mean to add code that would never be executed - so by throwing an error, it attempts to prevent you from inadvertently creating a code path that cannot be executed - instead forcing you to make a decision about it (even though, as you have proven, you still can work around it).

This error is mainly there to prevent programmer errors (a swap of 2 lines or more). In the second snippet, you make it clear that you don't care about the system.out.println().

Will it break the Java bytecode somehow, is it to protect the programmer or is it something else?
This is not required as far as Java/JVM is concerned. The sole purpose of this compilation error is to avoid silly programmer mistakes. Consider the following JavaScript code:
function f() {
return
{
answer: 42
}
}
This function returns undefined as the JavaScript engine adds semicolon at the end of the line and ignores dead-code (as it thinks). Java compiler is more clever and when it discoveres you are doing something clearly and obviously wrong, it won't let you do this. There is no way on earth you intended to have dead-code. This somehow fits into the Java premise of being a safe language.

http://docs.oracle.com/javase/specs/jls/se8/html/jls-14.html#jls-14.21
says:
14.21. Unreachable Statements
It is a compile-time error if a statement cannot be executed because it is unreachable.

Example 1:
In this case you are return before any statement because of it compiler never going to execute that code.
public static void main(String[] args) {
return;
System.out.println("unreachable");
}
In second code I have put the statement above of return and its work now :)
Example 2:
public static void main(String[] args) {
System.out.println("unreachable"); // Put the statement before return
return;
}
The reason behind this is that if you return sometime then code after it never going to execute because you already return the function data and as so it is shown unreachable code.

It's because it's a waste of resources for it to even be there. Also, the compiler designers don't want to assume what they can strip out, but would rather force you to remove either the code that makes it unreachable or the unreachable code itself. They don't know what is supposed to be there. There's a difference between the optimizations where they tweak your code to be a bit more efficient when it's compiled down to machine code and blatantly just removing code "you didn't need."

Does Java recognize infinite loops?

Given the following code sample:
public class WeirdStuff {
public static int doSomething() {
while(true);
}
public static void main(String[] args) {
doSomething();
}
}
This is a valid Java program, although the method doSomething() should return an int but never does. If you run it, it will end in an infinite loop. If you put the argument of the while loop in a separate variable (e.g. boolean bool = true) the compiler will tell you to return an int in this method.
So my question is: is this somewhere in the Java specification and are there situation where this behavior might be useful?

I'll just quote the Java Language Specification, as it's rather clear on this:
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.
...
A while statement can complete normally iff at least one of the following is true:
The while statement is reachable and the condition expression is not a constant expression with value true.
There is a reachable break statement that exits the while statement.
...
Every other statement S in a nonempty block that is not a switch block is reachable iff the statement preceding S can complete normally.
And then apply the above definitions to this:
If a method is declared to have a return type, then every return statement (§14.17) in its body must have an Expression. A compile-time error occurs if the body of the method can complete normally (§14.1).
In other words, a method with a return type must return only by using a return statement that provides a value return; it is not allowed to "drop off the end of its body."
Note that it is possible for a method to have a declared return type and yet contain no return statements. Here is one example:
class DizzyDean {
int pitch() { throw new RuntimeException("90 mph?!"); }
}

Java specification defines a concept called Unreachable statements. You are not allowed to have an unreachable statement in your code (it's a compile time error). A while(true); statement makes the following statements unreachable by definition. You are not even allowed to have a return statement after the while(true); statement in Java. Note that while Halting problem is undecidable in generic case, the definition of Unreachable Statement is more strict than just halting. It's deciding very specific cases where a program definitely does not halt. The compiler is theoretically not able to detect all infinite loops and unreachable statements but it has to detect specific cases defined in the spec.

If you are asking if infinite loops can be useful, the answer is yes. There are plenty of situations where you want something running forever, though the loop will usually be terminated at some point.
As to your question: "Can java recognized when a loop will be infinite?" The answer is that it is impossible for a computer to have an algorithm to determine if a program will run forever or not. Read about: Halting Problem
Reading a bit more, your question is also asking why the doSomething() function does not complain that it is not returning an int.
Interestingly the following source does NOT compile.
public class test {
public static int doSomething() {
//while(true);
boolean test=true;
while(test){
}
}
public static void main(String[] args) {
doSomething();
}
}
This indicates to me that, as the wiki page on the halting problem suggests, it is impossible for there to be an algorithm to determine if every problem will terminate, but this does not mean someone hasn't added the simple case:
while(true);
to the java spec. My example above is a little more complicated, so Java can't have it remembered as an infinite loop. Truely, this is a weird edge case, but it's there just to make things compile. Maybe someone will try other combinations.
EDIT: not an issue with unreachable code.
import java.util.*;
public class test {
public static int doSomething() {
//while(true);
while(true){
System.out.println("Hello");
}
}
public static void main(String[] args) {
doSomething();
}
}
The above works, so the while(true); isn't being ignored by the compiler as unreachable, otherwise it would throw a compile time error!

Yes, you can see these 'infinite' loops in some threads, for example server threads that listen on a certain port for incoming messages.

So my question is: is this somewhere in the Java specification
The program is legal Java according to the specification. The JLS (and Java compiler) recognize that the method cannot return, and therefore no return statement is required. Indeed, if you added a return statement after the loop, the Java compiler would give you a compilation error because the return statement would be unreachable code.
and are there situation where this behavior might be useful?
I don't think so, except possibly in obscure unit tests.
I occasionally write methods that will never return (normally), but putting the current thread into an uninterruptible infinite busy-loop rarely makes any sense.

After rereading the question....
Java understands while(true); can never actually complete, it does not trace the following code completely.
boolean moo = true;
while (moo);
Is this useful? Doubtful.

You might be implementing a general interface such that, even though the method may exit with a meaningful return value, your particular implementation is a useful infinite loop (for example, a network server) which never has a situation where it should exit, i.e. trigger whatever action returning a value means.
Also, regarding code like boolean x = true; while (x);, this will compile given a final modifier on x. I don't know offhand but I would imagine this is Java's choice of reasonable straightforward constant expression analysis (which needs to be defined straightforwardly since, due to this rejection of programs dependent on it, it is part of the language definition).

Some notes about unreachable statements:
In java2 specs the description of 'unreachable statement' could be found. Especially interesting the following sentence:
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.
So, it is not obviously possible to exit from while (true); infinite loop. However, there were two more options: change cached values or hack directly into class file or JVM operating memory space.