Why increment operator lambda compiles as Runnable?

Asked Jun 18 '17 at 14:49

Active Jun 19 '17 at 07:59

Viewed 133 times

Have a following code:

static int counter = 0;

static Callable<?> callable = () -> counter;

static Runnable runnable = () -> counter++;

static Runnable runnable2 = () -> counter; // DOES NOT COMPILE

public static void main(String[] args) throws Exception {
    runnable.run();
    System.out.println(callable.call()); // PRINTS "1"
}

I understand why runnable2 does not compile - because lambda returns a value and runnable does not have return type (is void). I also understand why callable compile, it returns counter variable value, which was incremented by calling run method. (I know it's single threaded but that's not the point).

What I don't understand is why runnable lambda actually compiles? As far as I know, ++ operator does return a value - incremented value - why does compiler ignore it?

asked Jun 18 '17 at 14:49

Gondy

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What if instead of `counter++` you had `method()`? What if `method` returns a value? Still compiles. – cbr Jun 18 '17 at 14:51
7

`counter` is an expression, `counter++` is both an expression and statement at the same time. – Willem Van Onsem Jun 18 '17 at 14:52
In the duplicate, if you follow the link for `ExpressionStatement`, you'll find the increment expression. – Sotirios Delimanolis Jun 18 '17 at 14:59
2

side-note: similar happens in a normal method: `counter;` is an error, `counter++;` is ok – user85421 Jun 19 '17 at 10:26

Why increment operator lambda compiles as Runnable?

0 Answers0