CallFunctionFast() question.
Posted: Mon May 06, 2024 6:01 am
When using CallFunctionFast() - what does it do behind the scenes?
the call is assembled to
I just assumed, no parameters, it would use the default specified in the parameter definition (ball.s="")
Code: Select all
Procedure who(ball.s="")
Debug "has the "+ball
EndProcedure
*p=@who()
CallFunctionFast(*p) ; not sending anything, so I expect the default parameter to to used ("").
CallFunctionFast(*p,@"") ; this works, but what if you don't know the default, or the default changes?
Code: Select all
integer r0=PB_CallFunctionFast(p0);