/**
* AST的根节点，解析的入口。
* @return
* @throws Exception
*/
private SimpleASTNode Prog(TokenReader tokens)
{
SimpleASTNode node = new SimpleASTNode(ASTNodeType.Programm, "pwc");
while (tokens.peek() != null)
{
SimpleASTNode child = intDeclare(tokens);
if (child == null)
{
child = expressionStatement(tokens);
}
if (child == null)
{
child = assignmentStatement(tokens);
}
if (child != null)
{
node.addChild(child);
}
else
{
throw new Exception("unknown statement");
}
}
return(node);
}
/**
* 乘法表达式
* @return
* @throws Exception
* M => int | int star M
*/
private SimpleASTNode multiplicative(TokenReader tokens)
{
SimpleASTNode child1 = primary(tokens); //消耗一个token
SimpleASTNode node = child1;
while (true) //循环调用 multiplicative规则
{
Token token = tokens.peek();
if (token != null && (token.getType() == TokenType.Star || token.getType() == TokenType.Slash))
{
token = tokens.read(); //消耗 * 或 /
SimpleASTNode child2 = primary(tokens);
if (child2 != null)
{
node = new SimpleASTNode(ASTNodeType.Multiplicative, token.getText());
node.addChild(child1);
node.addChild(child2);
child1 = node;
}
else
{
throw new Exception("invalid multiplicative expression, expecting the right part.");
}
}
else
{
break;
}
}
return(node);
}
/**
* 加法表达式
* @return
* @throws Exception
*
* 实现一个计算器，但计算的结合性是有问题的。因为它使用了下面的语法规则：
*
* additive -> multiplicative | multiplicative + additive
* multiplicative -> primary | primary * multiplicative //感谢@Void_seT，原来写成+号了，写错了。
*
* 递归项在右边，会自然的对应右结合。我们真正需要的是左结合。
*/
public SimpleASTNode additive(TokenReader tokens)
{
SimpleASTNode child1 = multiplicative(tokens); //应用add规则
SimpleASTNode node = child1;
if (child1 != null)
{
while (true)
{ //循环应用add'规则
Token token = tokens.peek();
if (token != null && (token.getType() == TokenType.Plus || token.getType() == TokenType.Minus))
{
token = tokens.read(); //读出加号
SimpleASTNode child2 = multiplicative(tokens); //计算下级节点
if (child2 != null)
{
node = new SimpleASTNode(ASTNodeType.Additive, token.getText());
node.addChild(child1); //注意，新节点在顶层，保证正确的结合性
node.addChild(child2);
child1 = node;
}
else
{
throw new Exception("invalid additive expression, expecting the right part.");
}
}
else
{
break;
}
}
}
return(node);
}
/**
* 语法解析：根节点
* @return
* @throws Exception
*/
public SimpleASTNode Prog(TokenReader tokens)
{
SimpleASTNode node = new SimpleASTNode(ASTNodeType.Programm, "Calculator");
SimpleParser parse = new SimpleParser();
SimpleASTNode child = parse.additive(tokens);
if (child != null)
{
node.addChild(child);
}
return(node);
}
/**
* 整型变量声明，如：
* int a;
* int b = 2*3;
*
* @return
* @throws Exception
*/
private SimpleASTNode intDeclare(TokenReader tokens)
{
SimpleASTNode node = null;
Token token = tokens.peek();
if (token != null && token.getType() == TokenType.Int)
{
token = tokens.read();
if (tokens.peek().getType() == TokenType.Identifier)
{
token = tokens.read();
node = new SimpleASTNode(ASTNodeType.IntDeclaration, token.getText());
token = tokens.peek();
if (token != null && token.getType() == TokenType.Assignment)
{
tokens.read(); //取出等号
SimpleASTNode child = additive(tokens);
if (child == null)
{
throw new Exception("invalide variable initialization, expecting an expression");
}
else
{
node.addChild(child);
}
}
}
else
{
throw new Exception("variable name expected");
}
if (node != null)
{
token = tokens.peek();
if (token != null && token.getType() == TokenType.SemiColon)
{
tokens.read();
}
else
{
throw new Exception("invalid statement, expecting semicolon");
}
}
}
return(node);
}
/**
* 表达式语句，即表达式后面跟个分号。
* @return
* @throws Exception
*/
private SimpleASTNode expressionStatement(TokenReader tokens)
{
int pos = tokens.getPosition();
SimpleASTNode node = additive(tokens);
if (node != null)
{
Token token = tokens.peek();
if (token != null && token.getType() == TokenType.SemiColon)
{
tokens.read();
}
else
{
node = null;
tokens.setPosition(pos); // 回溯
}
}
return(node); //直接返回子节点，简化了AST。
}
/**
* 基础表达式
* @return
* @throws Exception
*/
private SimpleASTNode primary(TokenReader tokens)
{
SimpleASTNode node = null;
Token token = tokens.peek();
if (token != null)
{
if (token.getType() == TokenType.IntLiteral) //int
{
token = tokens.read();
node = new SimpleASTNode(ASTNodeType.IntLiteral, token.getText());
}
else if (token.getType() == TokenType.Identifier)//变量
{
token = tokens.read();
node = new SimpleASTNode(ASTNodeType.Identifier, token.getText());
}
else if (token.getType() == TokenType.LeftParen) // (
{
tokens.read();
node = additive(tokens);
if (node != null)
{
token = tokens.peek();
if (token != null && token.getType() == TokenType.RightParen)
{
tokens.read(); //消耗右括号
}
else
{
throw new Exception("expecting right parenthesis");
}
}
else
{
throw new Exception("expecting an additive expression inside parenthesis");
}
}
}
return(node); //这个方法也做了AST的简化，就是不用构造一个primary节点，直接返回子节点。因为它只有一个子节点。
}
/**
* 赋值语句，如age = 10*2;
* @return
* @throws Exception
*/
private SimpleASTNode assignmentStatement(TokenReader tokens)
{
SimpleASTNode node = null;
Token token = tokens.peek(); //预读，看看下面是不是标识符
if (token != null && token.getType() == TokenType.Identifier)
{
token = tokens.read(); //读入标识符
node = new SimpleASTNode(ASTNodeType.AssignmentStmt, token.getText());
token = tokens.peek(); //预读，看看下面是不是等号
if (token != null && token.getType() == TokenType.Assignment)
{
tokens.read(); //取出等号
SimpleASTNode child = additive(tokens);
if (child == null)
{ //出错，等号右面没有一个合法的表达式
throw new Exception("invalide assignment statement, expecting an expression");
}
else
{
node.addChild(child); //添加子节点
token = tokens.peek(); //预读，看看后面是不是分号
if (token != null && token.getType() == TokenType.SemiColon)
{
tokens.read(); //消耗掉这个分号
}
else
{ //报错，缺少分号
throw new Exception("invalid statement, expecting semicolon");
}
}
}
else
{
tokens.unread(); //回溯，吐出之前消化掉的标识符
node = null;
}
}
return(node);
}
public void addChild(SimpleASTNode child)
{
children.Add(child);
child.parent = this;
}