[fricas-devel] E has (1:( ) -> E) does not work

[Prof. Dr. Johannes Grabmeier privat]

What is the proper way to qualify a type E to have a constant or a
function with no variables?
Other functions work well, here are examples from the algebra code.


if R has imaginary : () -> R then

if SMPF has _*: (NonNegativeInteger, SMPF) -> SMPF

if Coef has "*": (Expon,Coef) -> Coef then

Was not aware that both "*" and _* are allowed for infix functions.



Mit freundlichen Grüßen

Johannes Grabmeier

Prof. Dr. Johannes Grabmeier
Köckstraße 1, D-94469 Deggendorf
Tel. [phone], Tel. [phone]
Tel. [phone] (d), Fax: [phone]

[Bill Page]

Maybe like this:

(1) -> Integer has Zero:()->Integer

(1) true
Type: Boolean
(2) -> Integer has One:()->Integer

(2) true
Type: Boolean


As I recall 0 is just a pretty way to write Zero and 1 is just a
pretty way to write One.

On Thu, Jul 5, 2018 at 11:07 AM, Prof. Dr. Johannes Grabmeier privat

Post by Prof. Dr. Johannes Grabmeier privat
What is the proper way to qualify a type E to have a constant or a
function with no variables?
Other functions work well, here are examples from the algebra code.
if R has imaginary : () -> R then
if SMPF has _*: (NonNegativeInteger, SMPF) -> SMPF
if Coef has "*": (Expon,Coef) -> Coef then
Was not aware that both "*" and _* are allowed for infix functions.
Mit freundlichen Grüßen
Johannes Grabmeier
Prof. Dr. Johannes Grabmeier
Köckstraße 1, D-94469 Deggendorf
Tel. [phone], Tel. [phone]
Tel. [phone] (d), Fax: [phone]
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[Prof. Dr. Johannes Grabmeier FW]

yeah, I recall. But everywhere, we only have 0 and 1, so it is really
annoying always to fall into the same trap. Couldn't the compiler learn
about 0 and Zero etc.?

Post by Bill Page
(1) -> Integer has Zero:()->Integer
(1) true
Type: Boolean
(2) -> Integer has One:()->Integer
(2) true
Type: Boolean
As I recall 0 is just a pretty way to write Zero and 1 is just a
pretty way to write One.
On Thu, Jul 5, 2018 at 11:07 AM, Prof. Dr. Johannes Grabmeier privat

Post by Prof. Dr. Johannes Grabmeier privat
What is the proper way to qualify a type E to have a constant or a
function with no variables?
Other functions work well, here are examples from the algebra code.
if R has imaginary : () -> R then
if SMPF has _*: (NonNegativeInteger, SMPF) -> SMPF
if Coef has "*": (Expon,Coef) -> Coef then
Was not aware that both "*" and _* are allowed for infix functions.
Mit freundlichen Grüßen
Johannes Grabmeier
Prof. Dr. Johannes Grabmeier
Köckstraße 1, D-94469 Deggendorf
Tel. [phone], Tel. [phone]
Tel. [phone] (d), Fax: [phone]
--
You received this message because you are subscribed to the Google Groups "FriCAS - computer algebra system" group.
Visit this group at https://groups.google.com/group/fricas-devel.
For more options, visit https://groups.google.com/d/optout.

--
Mit freundlichen Grüßen

Johannes Grabmeier

Fraktionsvorsitzender
FREIE WÄHLER, Stadtrat Deggendorf

Prof. Dr. Johannes Grabmeier
Köckstraße 1, D-94469 Deggendorf
Tel. [phone], Tel. [phone]
Fax: [phone]