Package org.mvel2.util

Examples of org.mvel2.util.ExecutionStack.clear()

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    HashMap map = new HashMap();
    MapVariableResolverFactory mvrf = new MapVariableResolverFactory(map);

    for (int i = 0; i < 1000000; i++) {
      MVEL.executeExpression(s, mvrf);
      mvrf.clear();
    }

//        for (int x = 0; x < 4; x++) {
//            Serializable s = MVEL.compileSetExpression("tak.bar.name");
//            long time = System.currentTimeMillis();
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          case TERNARY:
            if (!stk.popBoolean()) {
              //noinspection StatementWithEmptyBody
              while (tk.nextASTNode != null && !(tk = tk.nextASTNode).isOperator(TERNARY_ELSE)) ;
            }
            stk.clear();
            continue;

          case TERNARY_ELSE:
            return stk.pop();
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             * If the program doesn't end here then we wipe anything off the stack that remains.
             * Althought it may seem like intuitive stack optimizations could be leveraged by
             * leaving hanging values on the stack,  trust me it's not a good idea.
             */
            if (tk.nextASTNode != null) {
              stk.clear();
            }

            continue;
        }

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            if ((Integer) stk.peek() == CHOR) {
              stk.pop();
              v1 = stk.pop();
              v2 = stk.pop();
              if (!isEmpty(v2) || !isEmpty(v1)) {
                stk.clear();
                stk.push(!isEmpty(v2) ? v2 : v1);
              }
              else stk.push(null);
            }
            else {
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          case TERNARY:
            if (!stk.popBoolean()) {
              //noinspection StatementWithEmptyBody
              while (tk.nextASTNode != null && !(tk = tk.nextASTNode).isOperator(TERNARY_ELSE)) ;
            }
            stk.clear();
            continue;

          case TERNARY_ELSE:
            return stk.pop();
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             * If the program doesn't end here then we wipe anything off the stack that remains.
             * Althought it may seem like intuitive stack optimizations could be leveraged by
             * leaving hanging values on the stack,  trust me it's not a good idea.
             */
            if (tk.nextASTNode != null) {
              stk.clear();
            }

            continue;
        }

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            if ((Integer) stk.peek() == CHOR) {
              stk.pop();
              v1 = stk.pop();
              v2 = stk.pop();
              if (!isEmpty(v2) || !isEmpty(v1)) {
                stk.clear();
                stk.push(!isEmpty(v2) ? v2 : v1);
              }
              else stk.push(null);
            }
            else {
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