[REBOL] Re: removing a character with replace - was: RE: [REBOL]
From: louisaturk:coxinet at: 6-Nov-2002 22:37
Anton,
At 12:50 PM 11/7/2002 +1100, you wrote:
>You need to analyze, Louis! :)
>Break the problem down.
>
> > I just tested again. At the console I get:
> >
> > >> x: ["aaaaaaa string"]
> > == ["aaaaaaa string"]
> > >> replace/all x "a" "s"
> > == ["aaaaaaa string"]
> > >> x
> > == ["aaaaaaa string"]
> > >> replace/all first "a" "s"
> > >> x: ["aaaaaaa string"]
> > == "sssssss string"
>
>I don't believe you. :)
>That should be
>
> replace/all first x "a" "s"
Sorry, I probably copied a mistyped line. Actually, I'm not sure what I
did. I stayed up too late trying to meet a deadline. It never pays
off. My brain starts slipping out of gear when I get too sleepy. :>)
>'replace takes three arguments.
>I will use parentheses to see clearly the
>first argument to replace:
>
> replace/all (first x) "a" "s"
>
>The two lines of code above are equivalent.
>If you want to see what first x is, use probe:
>
> replace/all (probe first x) "a" "s"
>
>or
>
> probe first x
> replace/all first x "a" "s"
>
> > However, this script doesn't work:
> >
> > rebol []
> > x: ["aaaaaaa string" "what is an apple good for"]
> >
> > foreach l x [
> > replace/all first l "a" "s"
> > ]
> >
> > ** Script Error: replace expected target argument of type: series
> > ** Near: replace/all first l "a" "s"
> >
> > But without first it works. What is happening?
> >
> > Louis
>
>foreach creates the nice situation where, each iteration
>through the loop, 'l is set to:
>
> first x
> second x
>
>so you already have what you want.
Ok, this is what was confusing me. I remember now reading that foreach
provided this feature.
>I recommend using ?? and 'probe again:
>
> foreach l x [
> ?? l ; <---- here you can see the value of 'l
> probe first l ; <---- let's just see what we would get...
> replace/all l "a" "s"
> ?? l ; <---- here is the result
> ]
>
>Anton.
I didn't know about ?? It will be useful for debugging.
Many thanks for explaining.
Louis
