Provided by: alliance_5.1.1-1.1_amd64 bug

NAME

       intersectbddnode - tests for an intersection between two bdd nodes.

SYNOPSYS

       #include "bdd101.h"
       bddnode ∗intersectbddnode( BddSystem, BddNode1, BddNode2 )
         bddsystem ∗BddSystem;
         bddnode   ∗BddNode1;
         bddnode   ∗BddNode2;

PARAMETERS

       BddSystem           The bdd system.

       BddNode1            The first bdd node.

       BddNode2            The second bdd node.

DESCRIPTION

       intersectbddnode  tests if the intersection of BddNode1 and BddNode2 exists, in the bdd system BddSystem.
       If a null pointer is given, the default bdd system is used.

RETURN VALUE

       intersectbddnode returns the bdd node zero  if  there  is  no  intersection,  and  a  computed  bdd  node
       otherwise.

EXAMPLE

       #include "bdd101.h"
          bddsystem  ∗BddSystem;
          bddcircuit ∗BddCircuit;
          bddnode    ∗BddNode1;
          bddnode    ∗BddNode2;
          bddnode    ∗BddImply;
          chain_list ∗Expr;
          BddSystem  = createbddsystem( 100, 1000, 100, 50000 );
          BddCircuit = createbddcircuit( "hello_world", 10, 10, BddSystem );
          Expr = createablbinexpr( ABL_AND,
                                   createablatom( "i0" ),
                                   createablatom( "i1" ) );
          BddNode1 = addbddcircuitabl( BddCircuit, Expr );
          freeablexpr( Expr );
          Expr = createablbinexpr( ABL_OR,
                                   createablatom( "i0" ),
                                   createablatom( "i1" ) );
          BddNode1 = addbddcircuitabl( BddCircuit, Expr );
          freeablexpr( Expr );
          BddImply = intersectbddnode( (bddsystem ∗)0, BddNode1, BddNode2 );
          Expr = convertbddcircuitabl( BddCircuit, BddNode );
          /* displays (i0 and i1)) */
          viewablexpr( Expr, ABL_VIEW_VHDL );
          freeablexpr( Expr );
          destroybddsystem( (bddsystem ∗)0 );
          destroybddcircuit( (bddcircuit ∗)0 );

SEE ALSO

       bdd(1)