FilterExpression.analyze

I'm puzzled by the following code fragment in this method:

        // if the filter is positional, try changing f[a and b] to f[a][b] to 
increase
        // the chances of finishing early.

        if (filterIsPositional &&
                filter instanceof BooleanExpression &&
                ((BooleanExpression)filter).operator == Tokenizer.AND) {
            BooleanExpression bf = (BooleanExpression)filter;
            if (isPositionalFilter(bf.operands[0]) &&
                    bf.operands[1].getItemType()==Type.BOOLEAN &&
                    !isPositionalFilter(bf.operands[1])) {
                FilterExpression f1 = new FilterExpression(start, 
bf.operands[0]);
                FilterExpression f2 = new FilterExpression(f1, bf.operands[1]);
                //System.err.println("Simplified to: ");
                //f2.display(10);
                return f2.analyze(env);
            }
            if (isPositionalFilter(bf.operands[1]) &&
                    bf.operands[1].getItemType()==Type.BOOLEAN &&
                    !isPositionalFilter(bf.operands[0])) {
                FilterExpression f1 = new FilterExpression(start, 
bf.operands[1]);
                FilterExpression f2 = new FilterExpression(f1, bf.operands[0]);
                return f2.analyze(env);
            }
        }

Translating the tests into words it reads:

If f is a positional filter of form "a and b" then
 If a is a positional filter and b is not a positional filter, but IS
  a BOOLEAN then rewrite

 Otherwise if b is a positional filter and a is not a positional
  filter, but B IS a BOOLEAN then rewrite
              ^
              |
I'm strugling to understand this, but the fact that the logic of the
two cases is not symmetrical (i.e. the same operand is testes for
being a boolean expression) worries me.
-- 
Colin Paul Adams
Preston Lancashire


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