MathAction #387 overloading functions in the interpreter does not work

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(1) -> *(A: Set INT, B: Set INT): Set INT == set concat [[(a*b)$Integer for a in members A] for b in members B]

   Function declaration ?*? : (Set(Integer), Set(Integer)) -> Set(
      Integer) has been added to workspace.

Type: Void

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2*3

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Compiling function * with type (Set(Integer), Set(Integer)) -> Set(
      Integer) 
   Conversion failed in the compiled user function * .

   Cannot convert the value from type Integer to Set(Integer) .

This problem is not related to *.

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mulset(A: Set INT, B: Set INT): Set INT == set concat [[(a*b)$Integer for a in members A] for b in members B]

   Function declaration mulset : (Set(Integer), Set(Integer)) -> Set(
      Integer) has been added to workspace.

Type: Void

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mulset(2,3)

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Compiling function mulset with type (Set(Integer), Set(Integer)) -> 
      Set(Integer) 
   Conversion failed in the compiled user function mulset .

   Cannot convert the value from type Integer to Set(Integer) .

It works if conversion from Integer to set is explicit.

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set [2] * set [3]

$\label{eq1}\left\{ 6 \right\}$

(1)

Type: Set(Integer)

overloading an operation in the interpreter is a problem --kratt6, Mon, 29 Oct 2007 08:53:07 -0600 reply

Why doesn't axiom find *: (INT, INT) -> INT anymore?

Agreed --Bill Page, Mon, 29 Oct 2007 09:08:51 -0600 reply

that is a problem. :-)

But you have overridden the default behaviour in the interpreter. It is ok if you are explicit:

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(2*3)$Integer

$\label{eq2}6$

(2)

Type: Integer

So I think probably the current behavior is sensible.

better title --kratt6, Mon, 29 Oct 2007 09:19:05 -0600 reply

Name: #387 redefining * => #387 overloading functions in the interpreter does not work

Subject: (replying) Be Bold !!

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