Test your SKILL Knowledge: Differences between similar operators

Try this:

brd=axlDBGetDesign()
allDrcs=brd->drcs
curDRCname=car(allDrcs) ->name
curDRCtype=car(allDrcs) ->type
curDRCxy=car(allDrcs) ->xy
printf("Drc '%s' of type '%s' is located at this coordinate: %L\n" curDRCname curDRCtype curDRCxy)

Or this:

brd=axlDBGetDesign()
allDrcs=brd->drcs
curDRCname=car(allDrcs ~>name)
curDRCtype=car(allDrcs ~>type)
curDRCxy=car(allDrcs ~>xy)
printf("Drc '%s' of type '%s' is located at this coordinate: %L\n" curDRCname curDRCtype curDRCxy)

maybe this:

;Find the name, type and xy of the first DRC in the design

curDRC = car(axlDBGetDesign()->drcs)

printf( "Drc %s type %s is located at this coordinate:%L\n" curDRC->name curDRC->type curDRC->xy)

Perfect!!Thanks eDave HJerry for replying. They are indeed very helpful and bring a very important aspect for all the SKILL programmers out there:

Difference between ~> and -> operators while extracting objects and attributes:

One of the main differences between these operators is that the ~> operator can handle a list on the left-hand side. For a single object these operators are more-or-less interchangeable, but if there is a list of database objects as input to the ~> operator (such as the result of another ~> query of an attribute that yields more than one value), then the ~> operator will apply the attribute query to each member of the list in turn and the result will be the list of each of these queries in turn.

Notice, curDRCname=car(allDrcs->name) this will give you nil as "allDrcs" is a list and "->" is not able to extract from it. On the other hand, this expression will yield correct results curDRCname=car(allDrcs ~>name)