Required Run Tools not accessible

[Bodekaer]

Hi @StephenHodgson ,
I've noticed that if I require specific tools/functions on a Run, which are not defined within Assistant.Tools, then the tool call cannot be found.
Is this intentional, or perhaps it could here default to searching any tool/function in the cache, and not only the specific assistant?

See e.g. the line here, where it fails to find the Run-only tool call:

com.openai.unity/OpenAI/Packages/com.openai.unity/Runtime/Assistants/AssistantExtensions.cs

Line 116 in 8b91c8c

var tool = assistant.Tools.FirstOrDefault(tool => tool.IsFunction && tool.Function.Name == toolCall.FunctionCall.Name) ??

[StephenHodgson]

@Bodekaer thanks for reporting. Can you give me a bit more context?

How exactly are you setting this up and how can I reproduce it?

[Bodekaer]

@StephenHodgson yes, here is a quick example to easily replicate it:

await _thread.CreateMessageAsync("Call TestingCustomRunTool tool");
run = await _thread.CreateRunAsync(new CreateRunRequest(_assistant, 
	tools: new [] { Tool.GetOrCreateTool(this, nameof(TestingCustomRunTool)) }, 
	toolChoice: "required")
);
run = await run.WaitForStatusChangeAsync();
while (run.RequiredAction != null)
{
	var toolOutputs = await _assistant.GetToolOutputsAsync(run.RequiredAction.SubmitToolOutputs.ToolCalls);
	run = await run.SubmitToolOutputsAsync(toolOutputs);
	run = await run.WaitForStatusChangeAsync();
}

[Bodekaer]

Update: It works fine now (without registering Run-only tools in Assistant Tools).

I simply:

1. Replaced assistant.Tools with with Tool.GetRegisteredTools() and
2. Added this to Tool.cs as well.
   public static IReadOnlyList<Tool> GetRegisteredTools() => toolCache;

[StephenHodgson]

There should be a more protected way to access to toolcache.

[Bodekaer]

Agree. Just temporary solution for now :)

[StephenHodgson]

not temporary, there is already the method for this:

Tool.GetAllAvailableTools(...);

[StephenHodgson]

I didn't want people to access the underlying collection directly bc it needs some initialization

[Bodekaer]

Ahh yes. I see now. Thanks. It works smoothly using that as well.