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🧩 Methods · Beginner

Return values & void in Java

return ends the method; every path must return in non-void methods.

🧩 The mysteryTwo if branches cover every number on Earth: s >= 50 and s < 50. The compiler still complains: "missing return statement". Is the compiler bad at math?

Handing back an answer

The header declares what a method gives back: int, String... **return value; hands that value to the caller and ends the method immediately**. Any code after it in that path doesn't run. The caller may use the result, or simply ignore it.

static int doubled(int x) {
    return x * 2;
}
int y = doubled(4); // y is 8
doubled(4);         // legal: result ignored
🔮 Predict it

Your turn

What prints?

static int check(int n) {
    if (n < 0) return -1;
    System.out.print("ok ");
    return n;
}
void main() {
    System.out.println(check(-5));
    System.out.println(check(3));
}
  1. ok -1 ok 3
  2. -1 ok 3
  3. -1 ok 3
Show the answer

For -5, return -1 ends the method before the print. For 3, the print runs (ok with no newline) and then 3 is returned, so println shows ok 3 on one line.

void: nothing to give

A **void method returns no value. It can still use a bare return; to exit early. But a void call is not a value**: println(log("hi")) won't compile ('void' type not allowed here). The reverse is also an error: a bare return; in an int method (missing return value).

static void log(String m) {
    if (m.isEmpty()) return; // exit early
    System.out.println(m);
}
static int bad() {
    return; // error: missing return value
}
⚠️ The trap

Every path must return

In a non-void method, every possible path must end in a return (or a throw). The compiler doesn't reason about your conditions: it sees if ... else if ... and imagines a path where neither is true, with no return. Use a plain else.

static String grade(int s) {
    if (s >= 50) {
        return "pass";
    } else if (s < 50) {
        return "fail";
    }
} // error: missing return statement
🔮 Predict it

Your turn

Does this compile and run?

static String sign(int n) {
    if (n > 0) {
        return "+";
    } else if (n <= 0) {
        return "-";
    }
}
void main() {
    System.out.println(sign(5));
}
  1. Prints +
  2. Prints -
  3. Compile error
Show the answer

Compile error: missing return statement. n > 0 and n <= 0 cover everything, but the compiler doesn't analyze conditions like that. Replace else if (n <= 0) with a plain else.

🤔 Think first

Return or throw

Can a non-void method end a path with throw instead of return?

Think about it, then reveal the answer

Yes. A path may end in a return or a throw. static int f() { throw new IllegalStateException(); } compiles: no path "falls off the end" without a value.

💼 In the real world

In real projects

A popular style is the guard clause: return early at the top of a method when the input is invalid (if (user == null) return;), so the main logic isn't buried inside nested ifs. Because return ends the method instantly, the rest can assume the input is fine.

Key takeaways

  1. return ends the method right away
  2. A void method may use a bare return; to exit early
  3. Every path through a non-void method must return a value
  4. The caller may ignore a returned value
🤯 Did you know?

static int f() { while (true) { } } compiles with no return at all! The loop can never finish, so there's no path that reaches the end without a value.

Practice questions

What does this print?

static int check(int n) {
    if (n > 10) return 1;
    System.out.print("small ");
    return 0;
}
void main() {
    System.out.println(check(20));
    System.out.println(check(5));
}
  1. small 1 small 0
  2. 1 0
  3. 1 small 0
  4. 1 small 0
Check your answer

1 small 0. For 20, return 1 ends the method before the print. For 5, the print runs and then 0 is returned, so println shows "small 0" on one line.

What does this print?

static void log(String m) {
    System.out.println(m);
}
void main() {
    System.out.println(log("hi"));
}
  1. hi
  2. hi null
  3. Compile error
  4. hi hi
Check your answer

Compile error. log returns void, which isn't a value, so it can't be passed to println: 'void' type not allowed here.

Next: when you pass your variable into a method, does the method get your variable, or a photocopy?