Does the position of a comment in a CompoundExpression matter for the result obtained with Out (%)?

Question

I would like to understand what happens when we include a comment at the end of a line that ends with a semicolon.

For example, if you execute

x;
%/.{x->1} (*some comment*);
%

you obtain 1. However, the following code:

x;
%/.{x->1};(*some comment*)
%

ignores the rule and the output is x.

The point is that if I remove the comment (*some comment*) from any of these two cases, the result is correct. So, what is the role of the position of the comment in all of this?

I do not know if it is a bug in my version (11.0) or if I have misunderstood something (related with the real meaning of ";" as a composition of expressions, or with the correct use of comments).

Answer 1

The trouble comes from the number of input lines used: If I copy your second example it just takes two lines - not three.

Note the difference:

1;
2;
%

2

1;
2; %

1

You will get a 1 in both of your examples if you press Return before the last %. % really is Out[] and will reference the previous line of input.

Edit

One should note that Out references to the result of a CompoundExpression (cf. the documentation).

In[1]:= CompoundExpression[2^2,Null] (* 2^2; *)
In[2]:= %

Out[2]= 4

But the behavior of Out is different for other expressions:

In[3]:= Null
In[4]:= %

(* No output will be shown, e.g. Out[4]= Null *)

The In- and Out-numbers here are simply exemplary.