This is a glitch likely to cause real problems
in a program expecting a scan derived function
to return its argument if empty.
Which should be the case.
Here everything is OK.
A„2 0 0 2½0
A(match)B„×™A ª A(match)×\[2]A ª A(match)×\[3]A ª A(match)×\A
1
1
1
1
0(match)†B
1
Here it is not anymore.
A„2 0 0 2½›¼3
A(match)B„×™A ª A(match)×\[2]A ª A(match)×\[3]A ª A(match)×\A
0
1
1
0
' '(match)†B
1