A hack using packages is given below. This currently works for T in {Integer,
Symbol}, List(T), Product(S,T) and Product(List(S),List(T)). Extension should
not be too painful...
Kindly, Mark.
-- base types
)abbrev package SHOW Show
Show() : Export == Implementation where
Export == with
show: Integer -> String
show: Symbol -> String
-- etc
Implementation == add
show (n:Integer) == string n
show (n:Symbol) == string n
-- etc
-- single type (e.g. homogeneous aggregates)
)abbrev package SHOW1 Show1
Show1(T:SetCategory) : Export == Implementation where
Export == with
if Show has (show:T->String) then
show: List(T) -> String
Implementation == add
if Show has (show:T->String) then
show(xs : List(T)) : String == -- print xs to string
empty? xs => "[]"
x := first xs
xs := rest xs
str := concat("[", show x)
while not empty?(xs) repeat
x := first xs
xs := rest xs
str := concat[str, ",", show x]
concat(str, "]")
-- two types
)abbrev package SHOW2 Show2
Show2(S:SetCategory, T:SetCategory) : Export == Implementation where
Export == with
if Show has (show:S->String) and Show has (show:T->String) then
show: Product(S,T) -> String
Implementation == add
if Show has (show:S->String) and Show has (show:T->String) then
show(p : Product(S,T)) : String ==
concat[ "(", show (selectfirst p), ",", show (selectsecond p), ")" ]
-- composition
)abbrev package SHOWN Shown
Shown(S:SetCategory, T:SetCategory) : Export == Implementation where
Export == with
if Show1(S) has (show:List(S)->String) and Show1(T) has
(show:List(T)->String) then
show: Product(List(S),List(T)) -> String
-- etc
Implementation == add
if Show1(S) has (show:List(S)->String) and Show1(T) has
(show:List(T)->String) then
show(p : Product(List(S),List(T))) : String ==
concat[ "(", show (selectfirst p), ",", show (selectsecond p), ")" ]
)if false
show 'a
show [1,2,3]
show makeprod(1,'a)$Product(Integer,Symbol)
show makeprod([1,2],[a,b])$Product(List(Integer),List(Symbol))
)end if
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