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"subsop" in "procedure" cannot work, weird!
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HI, all,
I found a weird problem:
When I used the subsop separately, it work fine. But when it was put in a procedure, then it cannot work. So weird!
For example:
>L:=[x,y,z,d];
L:=[x,y,z,d]
>L:=subsop(1=NULL,L);
L:=[y,z,d]
>L:=subsop(1=NULL,L);
L:=[z,d]
>L:=subsop(1=NULL,L);
L:=[d]
However, when I put the subsop in a procedure, it cannot work:
>F:=proc(L)
>L:=subsop(1=NULL,L);
>end proc;
>L:=[x,y,z,d];
L:=[x,y,z,d]
>F(L);
Error, (in F)illegal use of a formal parameter
----------------------------------------------------------------
Many thanks.
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assignment
It would be the same error with:
Without the assignment it works:
F:=proc(L) subsop(1=NULL,L); end proc; L:=[x,y,z,d]; F(L); [y, z, d]with side effects
If you really, really want to,
> F:=proc(L::evaln) > L:=subsop(1=NULL,eval(L)); > end proc: > L:=[x,y,z,d]; L := [x, y, z, d] > F(L); [y, z, d] > F(L); [z, d] > F(L); [d] > F(L); []Alternatively,
> restart: > F:=proc(L) > L:=subsop(1=NULL,eval(L)); > end proc: > L:=[x,y,z,d]; L := [x, y, z, d] > F('L'); [y, z, d] > F('L'); [z, d] > F('L'); [d] > F('L'); []Some people think that this sort of thing is evil.
acer
Thanks, acer, I find another
Thanks, acer,
I find another way to solve this problem. here is the way:
> F:=proc(L)
>local LL;
> LL:=subsop(1=NULL,LL);
>L:=LL;
> end proc:
> L:=[x,y,z,d];
L := [x, y, z, d]
>L:= F(L);
[y, z, d]
>L:= F(L);
[z, d]
>L:= F(L);
[d]
>F(L);
[]
By the way, shy it is evil? if I want to such thing, whether there are other methods that are not evil:)
Not that way
What you wrote won't work at all. Perhaps you meant something like
> F := proc(L) subsop(1=NULL, L) end proc;
L := [x,y,z,d];
L := F(L);
etc.
simpler
I would probably do it like this,
> F:=proc(L) > subsop(1=NULL,L); > end proc: > L:=[x,y,z,d]; L := [x, y, z, d] > L:= F(L); L := [y, z, d] > L:= F(L); L := [z, d] > L:= F(L); L := [d] > L:= F(L); L := []That keeps it as simple as possible, is easiest to understand, and is less confusing.
Doing it the other way, to bring about the change in L as a so-called side-effect on the name, can cause confusion. In a sense, it's deliberately subverting any protection Maple may otherwise give you about not having your procedure overwrite the formal parameter. As a general rule, you may not wish to do that unless it's necessary. For your example, the above way to overwrite L (explicitly, by the call to the proc) is more straightforward.
Way back in the day, before Maple allowed multiple assignments, some routines would use side-effects as a mechanism to fake multiple assignment. For example,
> f := proc(a,b) > b := 2*a; > 3*a; > end proc: > x:=11: > x := f(x,'y'): > x,y; 33, 22 > x := f(x,'y'): > x,y; 99, 66The above procedure changes both x and y. There was a time, long ago, when that was one of the easier ways to do get that effect. Nowadays, it can be done much more simply with a multiple assignment,
> restart: > f := proc(a) > 3*a, 2*a; > end proc: > x:=11: > x,y := f(x); x, y := 33, 22 > x,y := f(x); x, y := 99, 66When using side effects, one can get confused the by quotes, by either forgetting them or inserting them when they are not wanted.
> restart: > f := proc(a,b) > b := 2*a; > 3*a; > end proc: > x:=11: > x := f(11,y): # it works once, without quotes... > x,y; 33, 22 > x := f(11,y): # and now, the second time... Error, (in f) illegal use of a formal parameterAnd, doing it another way,
> f := proc(x::evaln) > x := 2*eval(x); > end proc: > x:=11: > f(x): > x; 22 > f('x'): Error, illegal use of an object as a nameAnd also, documenting its use for others is more involved if it's implemented in these ways.
acer
I agree
This simpler form is what I have shown above. I also dislike side effects. In fact, I think that the Maple library should be updated in this regard.
to acer: Yes, it is
to acer:
Yes, it is great.
Thanks so much.
to Robert Israel: yes,
to Robert Israel:
yes, when you want to newline, shift+enter instead of enter can get:
> F:=proc(L)
>local LL;
> LL:=subsop(1=NULL,LL);
> end proc:
> L:=[x,y,z,d];
L := [x, y, z, d]
>L:= F(L);
[y, z, d]
>L:= F(L);
[z, d]
>L:= F(L);
[d]
>F(L);
[]
actually, it is the same as:
F := proc(L) local LL; LL:=subsop(1=NULL,LL);(wihtout "enter")end proc;
Not about enter
That wasn't my point at all. Whether on one line or several, your code
will not work. The result of F(anything) is just NULL. Note that you haven't used L at all in the procedure.
The version you had before
won't work either, because the additional statement L := LL will generate an error " illegal use of a formal parameter".