On 12/22/2011 02:58 AM, Na, Hyunik wrote:
>
>
>> -----Original Message-----
>> From: jastadd-bounces_at_cs.lth.se [mailto:jastadd-bounces_at_cs.lth.se] On
>> Behalf Of Jesper Oqvist
>> Sent: Wednesday, December 21, 2011 8:38 PM
>> To: jastadd_at_cs.lth.se
>> Subject: Re: [Jastadd] [Q] unknown type ?
>>
>> Hello Hyunik,
>> I think the reason AssignExpr.sourceType() is the unkown type unless
>> it's a primitive type is to handle arithmetic assign expressions such as
>> AssignAddExpr with reference operands. The AssignSimpleExpr has the
>> following equation for sourceType:
>>
>> syn TypeDecl AssignSimpleExpr.sourceType() = getSource().type();
>>
>> which always returns the actual type of the source operand.
>>
> Aha! I missed this line.
>
> Given this, however, it still appears to me that the
> AssignSimpleExpr.typeCheck() should be as follows:
>
> public void AssignSimpleExpr.typeCheck() {
> if(!getDest().isVariable())
> error("left hand side is not a variable");
> else if ( sourceType().isUnknown() )
> error( "source type is unknown" );
> else if( !sourceType().assignConversionTo(getDest().type(),
> getSource()) )
> error( "can not assign " + getDest() + " of type " +
> getDest().type().typeName() +
> " a value of type " + sourceType().typeName() );
> }
>
> Am I missing something?
>
>
I have to agree that the above code seems more expressive than the
current implementation, but otherwise there is no problem with the
current implementation so I would suggest leaving it as it is.
If it's not broken, don't fix it! ;)
/Jesper
>
>> AssignSimpleExpr, I think, is the only assignment expression that works
>> with reference operands.
>>
>> /Jesper
>>
>> On 12/21/2011 10:36 AM, Na, Hyunik wrote:
>>
>>> Hello JastAddJ developers,
>>>
>>> In Java1.4Frontend/TypeCheck.jrag, the following line exists:
>>>
>>> syn TypeDecl AssignExpr.sourceType() =
>>> getSource().type().isPrimitive() ? getSource().type() : unknownType();
>>>
>>> which means the source type of an assign expression is the unknown type
>>> unless it is a primitive type.
>>> Why should it be?
>>>
>>> and the AssignSimpleExpr.typeCheck() is written as follows:
>>>
>>> public void AssignSimpleExpr.typeCheck() {
>>> if(!getDest().isVariable())
>>> error("left hand side is not a variable");
>>> else if(!sourceType().assignConversionTo(getDest().type(),
>>>
>> getSource())
>>
>>> && !sourceType().isUnknown())
>>> error("can not assign " + getDest() + " of type " +
>>> getDest().type().typeName() +
>>> " a value of type " + sourceType().typeName());
>>> }
>>>
>>> This means that, if the source type is unknown, it is not an error even
>>>
>> when
>>
>>> the source type is not assign-compatible
>>> with the destination type for a simple assignment.
>>> It is hard for me to figure out what's going on here.
>>>
>>> Why do you introduce the unknown type during the type checking process?
>>>
>>>
>>> - Hyunik
>>>
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>>>
>>>
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Received on Thu Dec 22 2011 - 16:12:41 CET