Overloading Operators

[Steinway Wu]

Hi, 

When I try to overload || with some_function, it says the || is not a symbol. Any idea?

Also, is it possible to overload + to be a unary postfix operator?

Thanks,

[Brandon Barker]

This may not be the solution you are looking for, but in ATS contrib, the file ./contrib/libfloats/TEST/test_LAgmat.dats has a macdef (the first one) that allows ^t to work as a postfix transpose operator.

Brandon Barker
brandon...@gmail.com

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[gmhwxi]

Only functions can be overloaded.

|| and && are already declared as macros in prelude/SATS/bool.sats, so
they cannot be overloaded. However, if you do

symintr &&
overload && with <some-function>

Then it works. But after this, && no longer refers to the original macro definition.

+ is current an infix operator. You need to do:

postfix +
overload + with <some-function>

[Yannick Duchêne]

Le mercredi 6 août 2014 20:05:28 UTC+2, gmhwxi a écrit :

Only functions can be overloaded.

But the function an operator symbol designates, may a polymorphic function… or it does work with polymorphic functions?

That's not the same as full overloading, this is just I feel this question is worth to be asked.

[Yannick Duchêne]

Le mercredi 6 août 2014 20:47:22 UTC+2, Yannick Duchêne a écrit :

But the function an operator symbol designates, may a polymorphic function… or it does work with polymorphic functions?

Sorry, typo. Please, read: “or it does *not* work with polymorphic functions?” 

[gmhwxi]

Yes, polymorphic functions can be overloaded.
Also, function templates can be overloaded as well.

[Steinway Wu]

Thanks! Although I don't think that's feasible since it overrides the logical operator. But this is a very clear answer.

[Yannick Duchêne]

What is the difference between this

infixl 10 +
overload + with foo

and this

infixl +
overload + with foo of 10

?

Do the `10` stands for the same precedence or do both means something different?

Le mercredi 6 août 2014 20:05:28 UTC+2, gmhwxi a écrit :

[gmhwxi]

>> infixl 10 +

10 is the precedence of '+'.

Suppose you have an expression '1 + 2 ++ 3';
if ++ has a higher precedence, then the expression
is parsed as 1 + (2 ++ 3); if ++ has a lower precedence,
you get (1 + 2) ++ 3.

>> overload + with foo of 10

Say you also have:

overload + with foo2 of 11

Suppose you have (x + y); if (x + y) can be translated to both foo(x, y) and foo2(x, y),
the foo2 is chosen (as it has a higher overloading value than foo).