ModuleOp generation from LambdaOp and FuncOp

[rbrchen]

Add support for generating a ModuleOp from either a LambdaOp or a FuncOp. The given LambdaOp will be converted into the root FuncOp, and the given FuncOp will be assigned as the root FuncOp.

For CoreOp.ModuleOp.ofLambdaOp(), a unique name for the new root FuncOp must be passed as a parameter.

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ModuleOp generation from LambdaOp and FuncOp

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[PaulSandoz]

You are traversing the model and resolving methods twice, you can combine the two. If you want to keep it a separate method then pass the temporary list of the encountered invokes to reflectable methods.

[PaulSandoz]

I find it easier to think in terms of pop/push. I see why you are doing this, to preserve the order of the call graph. Add a comment why you are working from the other end of the stack. You can accumulate into an array list (not a linked list) then do stack.addAll(temp.reversed()) (also choose a better name than temp).

Alternatively use pop and push, accumulating into a list and then pushing on the stack like so:

            for (FuncOp f : invokedFuncs.reversed()) {
                work.push(f);
            }

[PaulSandoz]

I don't think that check is right. To do this properly we would have to check against the names of methods encountered in the call graph. In general we have another naming clash problem with overloaded methods. A simple approach is to always postfix the function name with the topologically sorted index e.g. "_"

[PaulSandoz]

That's not making any assertions. Use the interpreter to invoke the function and lambda with the same arguments, and assert they produce the same result.