我在Roslyn中使用SyntaxRewriter遇到了棘手的情况。我想重写某些类型的语句,包括局部变量声明。该解决方案要求我将有问题的语句转换为多个语句,如下面的简单示例所示:
void method()
{
int i;
}
变为
void method()
{
int i;
Console.WriteLine("I declared a variable.");
}
我已经看过其他例子,其中块用于完成类似的事情,但当然在变量声明的情况下,声明的范围将受到影响。我提出了以下解决方案,但我正在喋喋不休。这似乎过于复杂,需要中断访客模式:
class Rewriter: SyntaxRewriter
{
public override SyntaxList<TNode> VisitList<TNode>(SyntaxList<TNode> list)
{
if (typeof(TNode) == typeof(StatementSyntax))
return Syntax.List<TNode>(list.SelectMany(st => RewriteStatementInList(st as StatementSyntax).Cast<TNode>()));
else
return base.VisitList<TNode>(list);
}
private IEnumerable<SyntaxNode> RewriteStatementInList(StatementSyntax node)
{
if (node is LocalDeclarationStatementSyntax)
return PerformRewrite((LocalDeclarationStatementSyntax)node);
//else if other cases (non-visitor)
return Visit(node).AsSingleEnumerableOf();
}
private IEnumerable<SyntaxNode> PerformRewrite(LocalDeclarationStatementSyntax orig)
{
yield return orig;
yield return Syntax.ParseStatement("Console.WriteLine(\"I declared a variable.\");");
}
}
我错过了什么?编辑语句并删除它们(通过空语句)似乎比重写多次更直接。
我接受了答案:
class Rewriter : SyntaxRewriter
{
readonly ListVisitor _visitor = new ListVisitor();
public override SyntaxList<TNode> VisitList<TNode>(SyntaxList<TNode> list)
{
var result = Syntax.List(list.SelectMany(_visitor.Visit).Cast<TNode>());
return base.VisitList(result);
}
private class ListVisitor : SyntaxVisitor<IEnumerable<SyntaxNode>>
{
protected override IEnumerable<SyntaxNode> DefaultVisit(SyntaxNode node)
{
yield return node;
}
protected override IEnumerable<SyntaxNode> VisitLocalDeclarationStatement(
LocalDeclarationStatementSyntax node)
{
yield return node;
yield return Syntax.ParseStatement("Console.WriteLine(\"I declared a variable.\");");
}
}
}
答案 0 :(得分:3)
我认为有一种简单的方法可以使Rewriter
更像访问者:使用其他访问者来处理列表中的节点:
class Rewriter: SyntaxRewriter
{
readonly Visitor m_visitor = new Visitor();
public override SyntaxList<TNode> VisitList<TNode>(SyntaxList<TNode> list)
{
var result = Syntax.List(list.SelectMany(m_visitor.Visit).Cast<TNode>());
return base.VisitList(result);
}
}
class Visitor : SyntaxVisitor<IEnumerable<SyntaxNode>>
{
protected override IEnumerable<SyntaxNode> DefaultVisit(SyntaxNode node)
{
return new[] { node };
}
protected override IEnumerable<SyntaxNode> VisitLocalDeclarationStatement(
LocalDeclarationStatementSyntax node)
{
return new SyntaxNode[]
{
node,
Syntax.ParseStatement(
"Console.WriteLine(\"I declared a variable.\");")
};
}
}
请注意,这不安全,如果从InvalidCastException
返回包含非TNode
对象的集合,则会抛出Visitor
。
答案 1 :(得分:2)
我不知道有一个更好的方法来处理这个问题。另一种稍微“访问者喜欢”的方法是使用VisitLocalDeclaration
来注释要替换的节点,例如:return (base.Visit(node).WithAdditionalAnnoations(myAnnotation);
。然后在VisitList中,您可以找到注释的子节点并在此时进行重写。
答案 2 :(得分:0)
我正在浏览Roslyn源代码,看看Roslyn团队自己是如何做到这一点的。这是一个例子:http://source.roslyn.codeplex.com/Microsoft.CodeAnalysis.CSharp.Features/R/bcd389b836bf7b4c.html
简而言之,我认为它看起来或多或少是这样的。 (这个重写器恰好删除了StatementExpressions,但你可以看到它是用迭代器方法构建的,所以添加方法也很容易。)
class TreeRewriter : CSharpSyntaxRewriter
{
public override SyntaxNode VisitBlock(BlockSyntax node)
=> node.WithStatements(VisitList(SyntaxFactory.List(ReplaceStatements(node.Statements))));
public override SyntaxNode VisitSwitchSection(SwitchSectionSyntax node)
=> node.WithStatements(VisitList(SyntaxFactory.List(ReplaceStatements(node.Statements))));
IEnumerable<StatementSyntax> ReplaceStatements(IEnumerable<StatementSyntax> statements)
{
foreach (var statement in statements)
{
if (statement is ExpressionStatementSyntax) continue;
yield return statement;
}
}
}
以下是我驾驶该代码的方式:
var rewriter = new TreeRewriter();
var syntaxTree = await document.GetSyntaxTreeAsync();
var root = await syntaxTree.GetRootAsync();
var newRoot = rewriter.Visit(root);
var newDocument = document.WithSyntaxRoot(newRoot);