CS606 Assignment No 04 Solution & Discussion Due Date:05-02-2013

[innocent larki]

Question No 1: Marks 20
Consider the following grammar productions. Suppose you have an attribute E.type which can
be set to either INT, BOOL. Assume that the type of an expression is set to INT if an error is
detected and you have a routine msg() similar to printf() that can be used to print error
message.
E → CONST {E.type=??}
| ID {E.type =getType(ID.name);}
| E1 +E2 {E.type=??}
| E1< E2 {E.type=??}
| E1= =E2 {E.type=??}
| (E1) {E.type=??}
You are required to add rules to the attribute grammar to calculate E.type for each grammar
production.

[salma iqbal]

also see handouts page 104...there is another possibility..

Handouts page frm where i took help are 109,and 113...check them..and inform me whether m right or wrong??..

[sny2ksa]

i hv pdf.

kindly tell topic names.

[salma iqbal]

Topic : Three-Address statement types...

check hint in it which is like a<b..

[sny2ksa]

ys chked...

u r wrong.

chk that attached idea solution in my previous comments.our type checking is right and

follow rules given lecture 38 under ur mentioned topic.

INt and BooL can't be conditional statements.

[salma iqbal]

ok..lemme make final solution..then will share..

[sny2ksa]

In fact " INT " and BOOL are types..we can't apply jumps on these.

[Golden Rock]

assignmnet ker li app ne mjuye se ni hui can u help me salma iqbal

[salma iqbal]

E ---->CONST             {E.type= INT ;}

    | ID                        {E.type =getType (ID.name) ;}

    | E1 +E2               {E.type=INT ;}

   | E1 +E2            {if ((E1.type==INT&& E2.type==INT))

                                E.type==INT;

                                else

                                E.type==BOOL;//error

                               }               

     | E1< E2              {E.type=BOOL ;}

    | E1< E2               {if ((E1.type== BOOL && E2.type== BOOL))

                                    E.type== BOOL;

                                    else

                                    E.type== INT; //error

                                    }   

    

     | E1= =E2          {E.type=BOOL ;}

     | E1= =E2          {if ((E1.type== BOOL && E2.type== BOOL))

                                    E.type== BOOL;

                                    else

                                    E.type== INT; //error

                                    }   

Still its incomplete solution..Confusion regarding ID is stil there and lso regarding      | (E1)...@sny2ksa..Is this correct solution??

[sny2ksa]

No,its not correct according to me...

still rules are missing in it..you are just printing errors.

if any one else can say about it..most welcome.

at least we will be able to submit it .

[waqas]

:-( koi kar day uplod .............sny2ksa ager koi mistak ha ti identify karo na plzzzzzz

[sny2ksa]

waqas.

assignment already has been discussed in the right way.

1st write expression types and then Rules.

see the last thread comments.you will get all with an idea solution.

[waqas]

smjh nai lag rai yaar :-(