Re: make RELEASE (here: kernel): template.ucl: line: 33, column: 0 - 'key must begin with a letter', character: '.'

Am Tage des Herren Sun, 5 Apr 2026 23:48:08 +0100
Lexi Winter <ivy@freebsd.org> schrieb:

> A FreeBSD User wrote in <20260405152059.6fa98991@thor.sb211.local>:
> > --- create-packages-kernel ---
> > /pool/sources/CURRENT/src/release/packages/generate-ucl.lua: fail to
> > parse(/pool/sources/CURRENT/src/release/packages/template.ucl): error while parsing
> > /pool/sources/CURRENT/src/release/packages/template.ucl: line: 33, column: 0 - 'key must
> > begin with a letter', character: '.' ***
> > [create-kernel-packages-extra-flavor-default-GENERIC] Error code 1  
> 
> this was broken by the libucl update (abda442d92fd) which disables
> macros in the Lua UCL parser by default.
> 
> please try this patch:
> 
> diff --git a/release/packages/generate-set-ucl.lua b/release/packages/generate-set-ucl.lua
> index 9ba7026b75df..7a1cddda7be1 100755
> --- a/release/packages/generate-set-ucl.lua
> +++ b/release/packages/generate-set-ucl.lua
> @@ -16,7 +16,7 @@ substituted as UCL variables.
>  local ucl = require("ucl")
> 
>  -- This parser is the output UCL we want to build.
> -local parser = ucl.parser()
> +local parser = ucl.parser(0)
> 
>  if #arg < 1 then
>         io.stderr:write(arg[0] .. ": missing template filename\n")
> diff --git a/release/packages/generate-ucl.lua b/release/packages/generate-ucl.lua
> index 19a9a95d5d38..aaaa4375dd4d 100755
> --- a/release/packages/generate-ucl.lua
> +++ b/release/packages/generate-ucl.lua
> @@ -124,7 +124,7 @@ local pkgprefix = nil
>  local pkgversion = nil
> 
>  -- This parser is the output UCL we want to build.
> -local parser = ucl.parser()
> +local parser = ucl.parser(0)
> 
>  -- Set any $VARIABLES from the command line in the parser.  This causes ucl to
>  -- automatically replace them when we load the source ucl.

Thank you very much! It worked - and I see, it has been already fixed in the tree, thanks for
this, too.
oh

-- 

A FreeBSD user