Re: raku-toml confusion

[email protected] (Peter Pentchev) Sat, 24 May 2025 12:40:35 +0300

Newsgroups	perl.perl6.users
Message-ID	<[email protected]>

On Fri, May 23, 2025 at 05:44:01PM -0700, ToddAndMargo via perl6-users wrote:
> Hi All,
> 
> https://github.com/tony-o/raku-toml
> 
> In the usage parsing example:
> 
>     Parsing TOML
>     use TOML;
>     my $config = from-toml("config.toml".IO.slurp);
>     # use $config like any ol' hash
> 
> This makes perfect sense to me.

...then you realize that the from-toml() function accepts a string,
not a file, right? The string that was read from the "config.toml"
file by the .IO.slurp() method.

> But in the Generating TOML example:
> 
>    use TOML;
>    my $config = {
>      bands => ['green day',
>                'motorhead',
>                't swift',],
>      favorite => 'little big',
>    };
>    my $toml-config = to-toml($config);
> 
> I am confused.  Seems to me he left something out.
> If this creates a TOML file for you, what/where
> is the file name and path?  What am I missing?

...and to-toml() also produces a string. If you want to write it to
a file, you need to do that yourself later.

Hope that helps!

G'luck,
Peter

-- 
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