On 2020-01-21 18:57, Tom Browder wrote:
On Tue, Jan 21, 2020 at 18:34 Todd Chester via perl6-users
<perl6-users@perl.org <mailto:perl6-users@perl.org>> wrote:
On 2020-01-21 16:09, Todd Chester via perl6-users wrote:
>> 4) A block (that is the { ... } bit) will always 'return' the last
>> expression evaluated.
>
> Seems to me I have see the last expression returned even without
> the {...}. Maybe I am misremembering.
Todd, the {} is the block defining the subroutine.
> sub AplusB( $a, $b --> Int ){$a+$b;}
The above is the sub's definition.
&AplusB
The above should generate an error because the mandatory args are
missing, depending on the context (don't quote me on that).
&AplusB is printer out by REPL. I don't know why. REPL
does that a lot.
> AplusB 2, 3
5
The above shows the sub being called with the required two args, and the
5 is the returned value which is probably shown in the REPL since there
is no semicolon after the call and the 5 is not usually seen otherwise.
$ p6 'sub AplusB( $a, $b --> Int ){$a+$b}; say AplusB 2, 3;'
5
-Tom
Hi Tom,
I was trying to cut things down to the simplest terms.
I was responding to Richard's statement
>> 4) A block (that is the { ... } bit) will always 'return' the last
>> expression evaluated.
I created a sub without the "{...}" and showed where it would
return the last equation without the "{...}". I may have
misunderstood Richard.
-T
