Order of output

Question

The code below:

def a
  print "Function 'a' called\n"
  99
end

print "a=", a, "\n"

produces:

Function 'a' called
a=99

Why does function 'a' called show first? I expected a= to be shown first.

Answer 1

Before arguments are passed to a method, they are evaluated (so that you have values to pass). Evaluation of a call to function a has a side effect of printing "function 'a' called. That's why it is printed first.

Answer 2

First, you define the method a; nothing is printed yet.

Then, when you get to the last line, the arguments to print are first evaluated before that statement prints anything. The first and last arguments are string literals. The middle argument is a call to the method a, which prints "Function 'a' called\n" before returning 99.

Then, the print statement that started all this is finally ready to print now that each of its arguments has been evaluated.