C++ BTreeNode::childOp方法代码示例

void Expression::compileConditional( const QString & expression, Code * ifCode, Code * elseCode )
{
	if( expression.contains(QRegExp("=>|=<|=!")) )
	{
		mistake( Microbe::InvalidComparison, expression );
		return;
	}
	if( expression.contains(QRegExp("[^=><!][=][^=]")))
	{
		mistake( Microbe::InvalidEquals );
		return;
	}
	// Make a tree to put the expression in.
	BTreeBase *tree = new BTreeBase();
	BTreeNode *root = new BTreeNode();

	// parse the expression into the tree
	buildTree(expression,tree,root,0);
	
	// Modify the tree so it is always at the top level of the form (kwoerpkwoep) == (qwopekqpowekp)
	if ( root->childOp() != equals &&
			root->childOp() != notequals &&
			root->childOp() != gt &&
			root->childOp() != lt &&
			root->childOp() != ge &&
			root->childOp() != le &&
			root->childOp() != pin &&
			root->childOp() != notpin &&
			root->childOp() != read_keypad )
	{
		BTreeNode *newRoot = new BTreeNode();
		
		BTreeNode *oneNode = new BTreeNode();
		oneNode->setChildOp(noop);
		oneNode->setType(number);
		oneNode->setValue("1");
		
		newRoot->setLeft(root);
		newRoot->setRight(oneNode);
		newRoot->setType(unset);
		newRoot->setChildOp(ge);
		
		tree->setRoot(newRoot);
		root = newRoot;
	}
	// compile the tree into assembly code
	tree->setRoot(root);
	tree->pruneTree(tree->root(),true);
	
	// We might have just a constant expression, in which case we can just always do if or else depending
	// on whether it is true or false.
	if( root->childOp() == noop )
	{
		if( root->value().toInt() == 0 )
			m_pic->mergeCode( elseCode );
		else
			m_pic->mergeCode( ifCode );
		return;
	}
	
	// traverse tree with argument conditionalRoot true
	// so that 3 == x gets integrated with code for if, repeat until etc...
	m_ifCode = ifCode;
	m_elseCode = elseCode;
	traverseTree(tree->root(),true);
	
	// Note deleting the tree deletes all nodes, so the root
	// doesn't need deleting separately.
	delete tree;
}

void BTreeBase::pruneTree(BTreeNode *root, bool /*conditionalRoot*/)
{
	Traverser t(root);
	
	t.descendLeftwardToTerminal();
	bool done = false;
	while(!done)
	{
	//t.descendLeftwardToTerminal();
	if( t.current()->parent() )
	{
		if( t.oppositeNode()->hasChildren() ) pruneTree(t.oppositeNode());
	}
	
	t.moveToParent();
	if( !t.current()->hasChildren() )
	{
		//if(t.current() == t.root()) done = true;
		if(!t.current()->parent()) done = true;
		continue;
	}

	BTreeNode *l = t.current()->left();
	BTreeNode *r = t.current()->right();
	BTreeNode *n = 0;
	BTreeNode *z = 0;
	

	// Deal with situations where there are two constants so we want
	// to evaluate at compile time
	if( (l->type() == number && r->type() == number) ) // && !(t.current()==root&&conditionalRoot) )
	{
		if(t.current()->childOp() == Expression::division && r->value() == "0" ) 
		{
			t.current()->setChildOp(Expression::divbyzero);
			return;
		}
		QString value = QString::number(Parser::doArithmetic(l->value().toInt(),r->value().toInt(),t.current()->childOp()));
		t.current()->deleteChildren();
		t.current()->setChildOp(Expression::noop);
		t.current()->setType(number);
		t.current()->setValue(value);
	}
	
	// Addition and subtraction
	else if(t.current()->childOp() == Expression::addition || t.current()->childOp() == Expression::subtraction)
	{
	// See if one of the nodes is 0, and set n to the node that actually has data,
	// z to the one containing zero.
	bool zero = false;
	if( l->value() == "0" )
	{
		zero = true;
		n = r;
		z = l;
	}
	else if( r->value() == "0" )
	{
		zero = true;
		n = l;
		z = r;
	}
	// Now get rid of the useless nodes
	if(zero)
	{
		BTreeNode *p = t.current(); // save in order to delete after

		replaceNode(p,n);
		t.setCurrent(n);
		// Delete the old nodes
		delete p;
		delete z;
	}
	}
	
	// Multiplication and division
	else if(t.current()->childOp() == Expression::multiplication || t.current()->childOp() == Expression::division)
	{
	// See if one of the nodes is 0, and set n to the node that actually has data,
	// z to the one containing zero.
	bool zero = false;
	bool one = false;
	if( l->value() == "1" )
	{
		one = true;
		n = r;
		z = l;
	}
	else if( r->value() == "1" )
	{
		one = true;
		n = l;
		z = r;
	}
	if( l->value() == "0" )
	{
		zero = true;
		n = r;
		z = l;
	}
//.........这里部分代码省略.........

void Expression::traverseTree( BTreeNode *root, bool conditionalRoot )
{
	Traverser t(root);
	t.start();
	
	// special case: if we are starting at the root node then
	// we are dealing with something of the form variable = 6
	// or variable = portb
	///TODO reimplement assignments as two branched trees?
	if ( t.current() == root &&
			!root->hasChildren() &&
			t.current()->childOp() != pin &&
			t.current()->childOp() != notpin &&
			t.current()->childOp() != function &&
			t.current()->childOp() != read_keypad )
	{
		switch(root->type())
		{
			case number: m_pic->assignNum(root->value()); break;
			case variable: m_pic->assignVar(root->value()); break;
			default: break; // Should never get here
		}
		// no need to traverse the tree as there is none.
		return;
	}
	
	t.setCurrent(root);
	
	if(t.current()->hasChildren())
	{
		// Here we work out what needs evaulating, and in which order.
		// To minimize register usage, if only one branch needs traversing,
		// then that branch should be done first.
		bool evaluateLeft = t.current()->left()->needsEvaluating();
	
		BTreeNode *evaluateFirst;
		BTreeNode *evaluateSecond;
	
		// If both need doing, then it really doesn't matter which we do
		// first (unless we are looking to do really complex optimizations...
	
		// Cases: 
		// - Both need evaluating,
		// - or left needs doing first,
		// in both cases we evaluate left, then right.
		if( evaluateLeft )
		{
			evaluateFirst = t.current()->left();
			evaluateSecond = t.current()->right();
		}
		// Otherwise it is best to evaluate right first for reasons given above.
		else
		{
			evaluateFirst = t.current()->right();
			evaluateSecond = t.current()->left();
		}
		
		QString dest1 = mb->dest();
		mb->incDest();
		QString dest2 = mb->dest();
		mb->decDest();
	
		bool evaluated = false;
		if( evaluateFirst->hasChildren() )
		{	
			traverseTree(evaluateFirst);
			evaluated = true;
		}
		else if( isUnaryOp(evaluateFirst->childOp()) )
		{
			doUnaryOp( evaluateFirst->childOp(), evaluateFirst );
			evaluated = true;
		}
		if ( evaluated )
		{
			// We need to save the result if we are going tro traverse the other
			// branch, or if we are performing a subtraction in which case the
			// value wanted in working is not the current value.
			// But as the optimizer will deal with unnecessary variables anyway,
			// always save to a register
			
			evaluateFirst->setReg( dest1 );
			evaluateFirst->setType( variable );
			m_pic->saveToReg( dest1 );
		}
	
		evaluated = false;
		if( evaluateSecond->hasChildren() )
		{
			mb->incDest();
			mb->incDest();
			traverseTree(evaluateSecond);
			evaluated = true;
			mb->decDest();
			mb->decDest();
		}
		else if( isUnaryOp(evaluateSecond->childOp()) )
		{
			doUnaryOp( evaluateSecond->childOp(), evaluateSecond );
			evaluated = true;
//.........这里部分代码省略.........