Hello,

This makes sense, but what if the variable is used again outside of the scope of the function?

function func a
	problem <- 3
	return a + problem
problem <- 5
print func(problem)

The above example would work with a result of 8, with problem being a variable inside and outside of the scope of func- in fact, even the following should work:

function func a
	problem <- 3
	return a + problem
print func(problem)

With a result of 3- as problem would be automatically instated as a variable with the value of 0:

variables do not need to be defined before being used in an expression, and are automatically initialised to the (real) value 0.0

So- am I correct in assuming that in both of the above scenarios, problem counts as 2 identifiers?