Re: Why does this work: ('5'):rep(4) --> 5555 ?

[Thomas Jericke <tjericke@...>]

On 09/11/2014 08:59 AM, Hao Wu wrote:

On Wednesday, September 10, 2014, Axel Kittenberger <axkibe@gmail.com> wrote:

The question should rather be, why do you need brackets for this?

I'd expect

'5':rep( 10 ).

to work. 

You need the parenthesis because

foo"bar":rep(5) will be translated to

(foo("bar")):rep(5)

 

On Wed, Sep 10, 2014 at 7:04 PM, Dirk Laurie <dirk.laurie@gmail.com> wrote:

2014-09-10 17:38 GMT+02:00 Robert McLay <mclay@tacc.utexas.edu>:
> Thanks all.  It is clear that this is safe to use.  At least until there is
> a good reason to remove support for it in later releases.
>
> I get that ("5",x,y) is going to return "5".  What I don't get is why then
> is:
>
>    s = '5':rep(4)
>
> does not work but
>
>    s = ('5'):rep(4)
>
> does.

prefixexp ::= var | functioncall | ‘(’ exp ‘)’

functioncall ::=  prefixexp args | prefixexp ‘:’ Name args

Since '5' is neither a variable nor a function call, it needs parentheses.

The paranthesis don't help, we actually had this discussion before:

foo()
("bar"):rep(5)

This code does not do what you might expect and it doesn't matter if you write ("bar"):rep(5) or "bar":rep(5)

The only thing that helps in this case is a semicolon after foo() to me clear what you want.

--
Thomas