Re: Discuss changing the behaviour of append()
- Posted by jimcbrown (admin) Jun 10, 2014
- 5040 views
jaygade said...
mattlewis said...
DerekParnell said...
object X, Y X = 4 Y = {8,9} X &= {Y} --Result is {4, {8,9}} X = {4} Y = {8,9} X &= {Y} --Result is {4, {8,9}} X = {4,5} Y = {8,9} X &= {Y} --Result is {4, 5, {8,9}} X = {} Y = {8,9} X &= {Y} --Result is {{8,9}}
So, if the change is made to Euphoria's append()/prepend() functions to do this, programs that now currently fail in this situation will no longer fail. Is that okay?
This seems like a positive change.
Matt
Maybe I'm not following very closely and misunderstanding. How in the first case would you concatenate to make a single-level sequence?
object X, Y X = 4 Y = {8,9} -- what operation do I need to form {4, 8, 9}? X = {4, 5} Y = {8,9} -- what operation do I need to form {4, 5, 8, 9}? X = "foo" Y = "bar" -- what operation do I need to form "foobar"?
It's like this:
object X, Y X = 4 Y = {8,9} X &= Y --Result is {4, 8, 9} X = 4 Y = {8,9} X &= {Y} --Result is {4, {8,9}} X = {4} Y = {8,9} X &= Y --Result is {4, 8, 9} X = {4} Y = {8,9} X &= {Y} --Result is {4, {8,9}} X = {4,5} Y = {8,9} X &= Y --Result is {4, 5, 8, 9} X = {4,5} Y = {8,9} X &= {Y} --Result is {4, 5, {8,9}} X = "foo" Y = "bar" X &= Y --Result is "foobar" X = "foo" Y = "bar" X &= {Y} --Result is NOT "foobar"
Basically, if you combine the concatenation operator with the new sequence operation, you've got a more expensive version of append.
