[b]Note:[/b] This does not remove arguments bound with [method bind].

Technically it does remove arguments bound with bind().

func _ready() -> void:
	# the 42 bound by `bind(42)` is unbound by `unbind(1)`.
	var f := foo.unbind(1).bind(42)
	# so you have to provide exactly one argument here.
	f.call(100)

func foo(a: int) -> void:
	print(a)

The key is that bind() and unbind() are decorators for the argument list provided to its own call() method, producing a new callable. So they are applied backwards when chained.

foo.unbind(1).bind(42).call(100)

call(100) #      argument list is: 100
bind(42)  #             append 42: 100, 42
unbind(1) # remove 1 from the end: 100

foo.bind(88, 99).unbind(1).call(22, 33)

call(22, 33) #      argument list is: 22, 33
unbind(1)    # remove 1 from the end: 22
bind(88, 99) #         append 88, 99: 22, 88, 99

Will add that it doesn't remove arguments previously bound for clarification

Clarified that description and added an example to illustrate it.

(My bad forgot func was a keyword so wanted to avoid it)

Thank you for your help

Thanks!

Thank you!

Cherry-picked for 4.0.3.