Why does scala call the method passed to println before printing the line it is called in?

Question

The program looks like this ...

object Delay{

    def main(args: Array[String]){
        delayed(time())
    }

    def time()={
        println(\"Getting time          


        

Answer 1

The reason why you see this execution order, is because t is a by-name parameter in your code:

def delayed(t: => Long)

If you defined your delayed method with a by-value parameter like so:

def delayed(t: Long)

the time() function would have been evaluated before the call to delayed, and you would get the following output instead:

Getting time in nanoseconds : 
In delayed Method
Param : 139735036142049

The trick is that by-name parameters are called only when they're used, and every time they're used. From the Scala docs:

By-name parameters are only evaluated when used. They are in contrast to by-value parameters.

By-name parameters have the advantage that they are not evaluated if they aren’t used in the function body. On the other hand, by-value parameters have the advantage that they are evaluated only once.