# [R] Break during the recursion?

Atte Tenkanen attenka at utu.fi
Sun Jul 15 16:13:54 CEST 2007

Oh. I forgot one extra-function:

pop.stack<-function(v){
if(length(v)==0){x=NA}
if(length(v)==1){x=v[1]; v=c()}
if(length(v)>1){x=v[1]; v=v[2:length(v)]}
return(list(vector=v,x=x))
}

Atte

> Hi,
>
> Is it possible to break using if-condition during the recursive
> function?
> Here is a function which almost works. It is for inorder-tree-walk.
>
> iotw<-function(v,i,Stack,Indexes) # input: a vector and the first
> index (1), Stack=c(), Indexes=c().
> {
> 	print(Indexes)
> 	# if (sum(i)==0) break # Doesn't work...
>
> 	if (is.na(v[i])==FALSE & is.null(unlist(v[i]))==FALSE)
>        	{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
> 	Indexes=c(Indexes,Stack[1])
> 	Stack=pop.stack(Stack)\$vector
> 	Indexes=c(Indexes,Stack[1])
> 	i=2*Stack[1]+1
> 	Stack=pop.stack(Stack)\$vector
> 	iotw(v,i,Stack,Indexes)
> }
>
>
> > v=c(`-`,`+`,1,`^`,`^`,NA,NA,"X",3,"X",2)
> > Stack=c()
> > Indexes=c()
>
> > iotw(v,1,Stack,Indexes)
> NULL
> NULL
> NULL
> NULL
> NULL
> [1] 8 4
> [1] 8 4
> [1] 8 4 9 2
> [1] 8 4 9 2
> [1] 8 4 9 2
> [1]  8  4  9  2 10  5
> [1]  8  4  9  2 10  5
> [1]  8  4  9  2 10  5 11  1
> [1]  8  4  9  2 10  5 11  1
> [1]  8  4  9  2 10  5 11  1  3
> Error in if (is.na(v[i]) == FALSE & is.null(unlist(v[i])) == FALSE)
> { :
> 	argument is of length zero
>
> Regards,
>
> Atte Tenkanen
> University of Turku, Finland
>