Defining function in js with $functions without extern fn silently does nothing

[timotheeguerin]

Add

export const $functions = {
  TestFn: {
    testFn: (context, target) => {
      return target;
    },
  },
};

then try to use with

op original(@typeChangedFrom(Versions.v2, int32) a: string, b: string): void;

alias T = testFn(original, #["b", "a"]);

@route("/test")
op test is T;

T is just nothing(not sure exactly but op test has no parmaeters), no errors.

I think we need to ban defining those functions

[github-actions]

🎯 Issue Triage

When a JS file exports $functions without a corresponding extern fn declaration in TypeSpec, the function silently produces incorrect results instead of raising an error. The suggestion is to ban or error on defining such functions without the matching extern declaration.

Category: Bug

Area: compiler:core

🔍 Analysis

• The $functions export mechanism allows JS-backed TypeSpec functions, but currently there's no validation requiring a matching extern fn declaration.
• When a function is defined via $functions without extern fn, calling it in TypeSpec produces no parameters on the resulting operation — silent incorrect behavior rather than a diagnostic.
• The fix likely involves the compiler's function resolution logic: when resolving a call to a function, it should verify that a corresponding extern fn exists, or alternatively, the $functions loader should reject entries that don't have a matching extern declaration.
• Affected area: compiler core, specifically the JS interop / extern function binding logic.

📝 Suggested next steps

• Add a compiler diagnostic (error or warning) when $functions exports a function that has no matching extern fn declaration in the TypeSpec type graph.
• Alternatively, validate at call-site that the function being called has a proper extern declaration.
• Add a test case covering this scenario to prevent regression.

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