Originally posted by: ShunjiTanaka

Hello.

I found that the following code does not work as I expected.

tuple TupA {
  int no;
  {int} set;
};
tuple TupB {
  int no;
  {string} set;
};

{TupA} A = { <0, {0}>,   <1, {0}>,   <2, {0}>,   <3, {1}>,   <4, {1}> };
{TupB} B = { <0, {"0"}>, <1, {"0"}>, <2, {"0"}>, <3, {"1"}>, <4, {"1"}> };

tuple Tup {
  int a;
  int b;
};

// It works.
{Tup} x = {<a.no, b.no> | ordered a, b in A : item(a.set, 0) == item(b.set, 0)};
// It does not work.
{Tup} y = {<a.no, b.no> | ordered a, b in B : item(a.set, 0) == item(b.set, 0)};

// A simple workaround is to introduce another tuple.
tuple TupC {
  int a_no;
  int b_no;
  string a_string;
  string b_string;
};

{TupC} C = {<a.no, b.no, item(a.set, 0), item(b.set, 0)> | ordered a, b in B};
{Tup} z = {<u.a_no, u.b_no> | u in C : u.a_string == u.b_string};

execute {
  writeln(x);
  writeln(y);
  writeln(z);
}

The three sets, x, y, and z should be the same, but y is {<0 1> <0 2> <3 4>} instead of {<0 1> <0 2> <1 2> <3 4>}.

This happens when elements of string sets in a tuple are compared by item().  I would like to know if it is a specification of OPL or not.

I tested the code by:

• platforms: Linux x86-64 and Windows 7 x86-64
• versions: 12.6.0 and 12.6.1
• executables: oplide and oplrun

Regards,

Shunji

Hi,

indeed there seems to be an issue.

I would write:

{Tup} y = {<a.no, b.no> | ordered a, b in B : 1==card(a.set inter b.set) };
 

instead of your workaround.

You may notice that if you rewrite:

// It works.
{Tup} x = {<a.no, b.no> | ordered a, b in A : first(a.set) == first(b.set)};
// It does not work.
{Tup} y = {<a.no, b.no> | ordered a, b in B : first(a.set) == first(b.set)};

then x works and y gives a concert error

regards

Originally posted by: ShunjiTanaka

Thank you for your reply.

I understand it is an issue.  I'll wait for it fixed.

Your workaround is perfect when the number of elements in the set is always one.  However, as you know, we need not use a set in such a case. ;-)

The following workaround will be more memory efficient than my first one:

tuple TupD {
  int no;
  string s;
};

{TupD} D = {<a.no, first(a.set)> | a in B};
{Tup} u = {<a.no, b.no> | ordered a, b in D : a.s == b.s};

then x works and y gives a concert error

Yes.  I first tried first(), but I gave it up due to the error.  I had completely forgotten it till your reply.

Regards,

Shunji