SKILL Lists

Hi Leonardo,

You are doing two things wrong:

1. if you do:

List_Original='(
                          ((0 1)  (2 3))
                         )

and

List_A=List_Original

then List_A and List_Original will point to the same object, if you modify one it means you modify the other one

You will have to store a copy version in order to do whatever suits you:

List_A = (copy List_Original)

2. you cannot use = with a function call on the left side:

you can use setf to achieve what you want:

(setf cadr(nth(0 val))  cadr(nth(0 val)) + 0.5)

should work

I really don't like the combined use of both cadr and nth (those are two ways of doing the same thing, I think you should stick to one)

In your case, you can use topEdge (bottomEdge, leftEdge or rightEdge works as well)

Your final code should look like this:

List_A = (copy List_Original)

(foreach bbox List_A

  (setf (bottomEdge bbox) (bottomEdge bbox)+.5)

  (setf (topEdge bbox)       (topEdge bbox)+.5)

  )

Cheers,

Aurel

Hi Leonardo,

You are doing two things wrong:

1. if you do:

List_Original='(
                          ((0 1)  (2 3))
                         )

and

List_A=List_Original

then List_A and List_Original will point to the same object, if you modify one it means you modify the other one

You will have to store a copy version in order to do whatever suits you:

List_A = (copy List_Original)

2. you cannot use = with a function call on the left side:

you can use setf to achieve what you want:

(setf cadr(nth(0 val))  cadr(nth(0 val)) + 0.5)

should work

I really don't like the combined use of both cadr and nth (those are two ways of doing the same thing, I think you should stick to one)

In your case, you can use topEdge (bottomEdge, leftEdge or rightEdge works as well)

Your final code should look like this:

List_A = (copy List_Original)

(foreach bbox List_A

  (setf (bottomEdge bbox) (bottomEdge bbox)+.5)

  (setf (topEdge bbox)       (topEdge bbox)+.5)

  )

Cheers,

Aurel

I would do it this way, which is much more immune to side effects:

List_A=foreach(mapcar bBox List_Original
    list(
        xCoord(lowerLeft(bBox)):yCoord(lowerLeft(bBox))+0.5
        xCoord(upperRight(bBox)):yCoord(upperRight(bBox))-0.5
    )
)

Unfortunately Aurel, your approach still destructively modifies the original list. That's because copy() doesn't do a deep copy - it only generates a new top-level list, and your destructive operations are one level down.

Regards,

Andrew

My bad,

you're definitely right, the foreach mapcar approach is much more robust and natural (even though the foreach mappings might not be obvious for users experienced with other languages)

Thanks for your rectification

Cheers,

Aurel

Hi Aurel, and  Andrew Beckett

= is an assignment operator, so the value of variable on RHS is assigned to the value of variable in LHS. Then why after I change the variable on LHS, which is List_A, when i do the foreach, is not reflected? Since the code runs serially, the latest foreach loop should update the values of List_A right?  

Please correct me.

Thanks,

Leo.

Leo,

You can assign to a variable, but you cannot assign to the result of a function - that is not a variable. So when you were doing:

cadr(nth(0 val))=a

that will not work. You could use setf() to do this kind of assignment, but then you'd be modifying the original list, which you explicitly said you don't want to do.

Andrew