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🔀 Control Flow · Beginner

Infinite loops & unreachable code in Java

while(true), compiler unreachable-statement errors.

🧩 The mysteryYou write a perfectly innocent println after a loop, and Java refuses to compile it, calling it "unreachable". How can the compiler predict the future?

Forever, on purpose

while (true) (or for (;;)) loops forever unless a break, return or exception gets out. That's not a bug: game loops, servers and input loops all work this way, with an exit inside.

while (true) {
    String cmd = readCommand();
    if (cmd.equals("quit")) break;
    run(cmd);
}
System.out.println("bye");
🔮 Predict it

Your turn

What prints?

int n = 3;
while (true) {
    n = n * 3;
    if (n > 50) break;
}
System.out.println(n);
  1. 27
  2. 81
  3. 50
  4. Compile error
Show the answer

n triples: 9, 27, 81. Only at 81 is n > 50, so break exits and 81 prints. The println is reachable because the loop contains a break.

The compiler does the math

If a while (true) has no break, it can never finish normally, so any statement after it can never run. Java makes that a compile error: unreachable statement. The check works with constants like true; the compiler doesn't evaluate your variables.

int x = 0;
while (true) {
    x++;
}
System.out.println(x); // unreachable
🔮 Predict it

Your turn

What happens?

for (;;) {
    System.out.println("tick");
}
System.out.println("done");
  1. Prints tick forever
  2. Compile error
  3. Prints done
Show the answer

Compile error. for (;;) has no condition and the body has no break, so the compiler proves done can never print, and rejects the program before it ever runs.

⚠️ The trap

Dead code after a jump

A statement placed directly after return, break or continue in the same block can never run. That's also an unreachable statement compile error.

for (int i = 0; i < 3; i++) {
    continue;
    System.out.println(i); // error
}
🤔 Think first

A curious exception

if (false) { ... } compiles fine, but while (false) { ... } doesn't. Why the difference?

Think about it, then reveal the answer

The language spec special-cases if so you can switch code off with a constant flag like if (DEBUG). A loop whose condition is the constant false has an unreachable body, which is an error. Same for for (;false;).

💼 In the real world

In real projects

Event loops in servers, games and GUI frameworks are while (true) loops at heart, each with a clear exit. The unreachable-statement check also earns its keep during refactoring: it flags leftover code after an early return that someone forgot to delete.

Key takeaways

  1. while (true) { ... if (done) break; } is a normal, legal pattern
  2. A statement after an infinite loop with no break doesn't compile
  3. Code directly after return, break or continue in the same block is unreachable
  4. if (false) { ... } is allowed on purpose; while (false) { ... } is not
🤯 Did you know?

In 1936 Alan Turing proved that no program can decide, for every program, whether it will halt. That's why javac only checks simple cases like while (true) and can't catch every infinite loop.

Practice questions

What does this print?

int n = 1;
while (true) {
    n *= 2;
    if (n > 20) break;
}
System.out.println(n);
  1. 16
  2. 32
  3. 20
  4. Compile error
Check your answer

32. n doubles: 2, 4, 8, 16, 32. Only at 32 is n > 20, so break exits and 32 prints. The println is reachable thanks to the break.

What does this print?

int x = 0;
while (true) {
    x++;
}
System.out.println(x);
  1. Prints 0
  2. Compile error
  3. Runs forever, prints nothing
  4. Throws StackOverflowError
Check your answer

Compile error. The loop has no break, so the compiler knows it can never finish normally. The println after it is an unreachable statement.

Next: stop hard-coding values and let the user type them. But beware the Enter key that eats your next input.