Provided by: alliance_5.1.1-1.1build1_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)