How can I get an exact BBox from lower level to top?

Hi Jungyoon,

Something like this,

cb = centerBox(css()~>bBox) ; shape's centerbox inside instance.
 inst_transform = css()~>transform ;instance in the top level
 location = dbTransformPoint(cb inst_transform) ;location of instance's shape to the top level.

You have to modify it to get the bBox, because the code gets the center box location.

Best regards,

Marben

Thank you for replying Marben,

What you mean is dbTransformPoint works for mosaic as well unlike dbGetHierPathTransform?

Hi Jungyoon,

I haven't try it in mosaic, because I'm busy now.

You should try it and message me if it works or not.

I will fix it if it will not work.

Best regards,

Marben

Hi Jungyoon,

Not sure Marben's reply helps particularly - the complexity is that you need to know which row and column of the mosaic you are expecting to transform. dbGetHierPathTransform handles that already though - provided you give the input appropriately.

If you use dbGetOverlaps or dbGetTrueOverlaps there's a fifth argument called doRowCol which adds the row and column information. For example:

dbGetTrueOverlaps(geGetEditCellView() list(9:8.1 9:8.1) "Metal3" 3)
((db:0x234df29b db:0x234df41a))
dbGetTrueOverlaps(geGetEditCellView() list(9:8.1 9:8.1) "Metal3" 3 t)
(((db:0x234df29b 0 1) db:0x234df41a))

Notice the extra information - the first entry is now a list of the mosaic and row and column. So what I would do is use my abGetHierPathShape function (probably on this forum somewhere, but just in case):

/***************************************************************
*                                                              *
*                 abGetHierPathShape (overlap)                 *
*                                                              *
*             Find the leaf shape from an overlap              *
*                                                              *
***************************************************************/

(defun abGetHierPathShape (overlap)
  (if (listp overlap)
    (abGetHierPathShape (cadr overlap))
    overlap))

then do:

overlaps=dbGetTrueOverlaps(geGetEditCellView() list(9:8.1 9:8.1) "Metal3" 3 t)
foreach(overlap overlaps
  shape=abGetHierPathShape(overlap)
  transform=dbGetHierPathTransform(overlap)
  printf("transformed bbox is %L\n" dbTransformBBox(shape~>bBox transform))
)

If you're not using dbGetOverlaps or dbGetTrueOverlaps, you would need to construct a similar list representing the hierarchical path to the shape and pass that to dbGetHierPathTransform.

Regards,

Andrew.

Thank you so much for your advice, Marben and Andrew.

Now I remember the case. I tried dbTransformPoint to get a coordinate but it didn't work for mosaic. So I am testing the function abGetHierPathShape now.

But I cannot understand the code fully so I'd like to explain what I understand on it. It looks like abGetHierPathShape function gives me "shape = overlap" when I wrote shape = abGetHierPathShape(overlap) because it is returning overlap value. Can you please explain the function of abGetHierPathShape and what I understand wrongly, Andrew?

If I have more mosaics in a hierarchy to get down the shape, I am wondering if this function would work or not.   

Again, you've not explained how you are getting the input to the function - are you using dbGetOverlaps or dbGetTrueOverlaps? The abGetHierPathShape processes one of the "overlap" descriptions returned by these functions, and gives you the leaf shape. It works with mosaics, as does dbGetHierPathTransform. The example below was a mosaic of an instance of a mosaic:

overlaps=dbGetOverlaps(geGetEditCellView() list(6.5:-6 6.5:-6) "Metal3" 32 t)
(((db:0x24e0f39a 1 0)
     (db:0x24e0f71a
          ((db:0x24e0fb9a 1 2) db:0x24e0fe9b)
     )
     )
)
firstOverlap=car(overlaps)
((db:0x24e0f39a 1 0)
     (db:0x24e0f71a
     ((db:0x24e0fb9a 1 2) db:0x24e0fe9b)
     )
)

shape=abGetHierPathShape(firstOverlap)
db:0x24e0fe9b
transform=dbGetHierPathTransform(firstOverlap)
((3.445 -8.73) "MXR90" 1.0)
dbTransformBBox(shape~>bBox transform)
((6.125 -6.635)
     (8.565 -4.595)
)

I've put in bold everything I typed, and the other text is the outputs shown in the CIW. I've cross-highlighted (in the same colour) what abGetHierPathShape returns.

The situation where abGetHierPathShape returns the same as its input is if you just pass it a database object directly - i.e. not one of these lists describing an overlap. So if it was a top-level shape, the leaf shape is the shape itself. The function recursively follows the list of lists down to find the leaf shape, but only does that if the input was a list in the first place.

Regards,

Andrew.

Andrew,

Thank you so much!! I suddenly had an urgent tapeout last week so I’m late to test it. This issue was a problem I have not solved so far.

It worked very well. You are such an amazing guy. Thanks again. 

Jungyoon