Fix React.forwardRef parameter handling in JSX v4 transformation

[Copilot]

The React.forwardRef transformation was incorrectly treating the ref parameter as an optional argument instead of a positional argument, causing compilation errors in v12.

Problem

React.forwardRef functions were failing to compile with the error:

This function call is incorrect.
  The function has type:
  (
  props<string, React.element, ReactDOM.Ref.currentDomRef>,
  Js.Nullable.t<ReactDOM.Ref.currentDomRef>,
) => Jsx.element
  
  It is called with 2 argument just but requiress 1.

This affected code like:

@react.component
let make = React.forwardRef((~className=?, ~children, ref) =>
  <div>
    <input
      type_="text"
      ?className
      ref=?{ref->Nullable.toOption->Option.map(ReactDOM.Ref.domRef)}
    />
    children
  </div>
)

Root Cause

In the JSX v4 transformation, the ref parameter was being processed as Optional {txt = "ref"} instead of Nolabel. This created a mismatch where:

• The function signature expected: (props, ~ref=?) => ...
• But the generated code called it as: fn(props, ref)

Solution

Modified recursively_transform_named_args_for_make in jsx_v4.ml to treat the ref parameter as a positional argument (Nolabel) instead of optional (Optional). This ensures the function signature matches the calling convention:

• Function signature: (props, ref) => ...
• Function call: fn(props, ref) ✓

The fix aligns with the JSX V4 specification where React.forwardRef should generate:

React.forwardRef({
  let fn = @react.component (~x, ~y) => ref => body
  (props, ref) => fn(props, ref)
})

Fixes #7738.

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[cristianoc]

nonsense