本文整理汇总了C#中Microsoft.CodeAnalysis.SyntaxTree.TryGetRoot方法的典型用法代码示例。如果您正苦于以下问题:C# SyntaxTree.TryGetRoot方法的具体用法?C# SyntaxTree.TryGetRoot怎么用?C# SyntaxTree.TryGetRoot使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Microsoft.CodeAnalysis.SyntaxTree
的用法示例。
在下文中一共展示了SyntaxTree.TryGetRoot方法的3个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: GetMatchingSyntaxNodes
public IEnumerable<SyntaxNode> GetMatchingSyntaxNodes(SyntaxTree syntaxTree)
{
SyntaxNode root;
if (syntaxTree.TryGetRoot(out root))
{
// Just return the root if not filtering by declaration
if (string.IsNullOrWhiteSpace(_declarationFilter))
{
return new[] { root };
}
else
{
// Filter by declaration
return root
.DescendantNodes(x => !SyntaxNodeMatchesDeclaration(x))
.Where(x => SyntaxNodeMatchesDeclaration(x));
}
}
return new SyntaxNode[] { };
}
示例2: GetAnyCallNodeAboveTriggerPoint
public static SyntaxNode GetAnyCallNodeAboveTriggerPoint(ITrackingPoint triggerPoint, ITextSnapshot snapshot, SyntaxTree parseTree) {
Contract.Requires(snapshot != null);
Contract.Requires(parseTree != null);
Contract.Requires(triggerPoint != null);
SyntaxNode syntaxRoot;
if (!parseTree.TryGetRoot(out syntaxRoot))
return null;
var leafNode = syntaxRoot.FindToken(triggerPoint.GetPosition(snapshot), false);
if (leafNode.IsKind(SyntaxKind.None))
return null;
//Is anyone in our ancestry a call node?
var nodeInQuestion = leafNode.Parent;
SyntaxNode ptn = null;
while (nodeInQuestion != null) {
//Is the node in question a node call?
var asCall = nodeInQuestion as InvocationExpressionSyntax;
if (asCall != null) {
ptn = asCall;
break;
}
var asCtorCall = nodeInQuestion as ObjectCreationExpressionSyntax;
if (asCtorCall != null) {
ptn = asCtorCall;
break;
}
//Climb higher up our ancestry for the next iteration
nodeInQuestion = nodeInQuestion.Parent;
}
//Did we successfully find a call node?
return ptn;
}
示例3: GetTargetAtTriggerPoint
public static SyntaxNode GetTargetAtTriggerPoint(ITrackingPoint triggerPoint, ITextSnapshot snapshot, SyntaxTree parseTree)
{
Contract.Requires(snapshot != null);
Contract.Requires(parseTree != null);
Contract.Requires(triggerPoint != null);
SyntaxNode syntaxRoot;
if (!parseTree.TryGetRoot(out syntaxRoot))
return null;
var leafToken = syntaxRoot.FindToken(triggerPoint.GetPosition(snapshot), false);
if (leafToken.IsKind(SyntaxKind.None))
return null;
var leafNode = leafToken.Parent;
var asPropDecl = leafNode as PropertyDeclarationSyntax;
if (asPropDecl != null)
{
return asPropDecl;
}
//Is our leaf node a name?
var asName = leafNode as IdentifierNameSyntax;
if (asName == null)
return null;
//Is anyone in our ancestry a call node?
var nodeInQuestion = leafNode;
var lastNonBinaryOperatorNode = leafNode;
var lastNode = leafNode;
while (nodeInQuestion != null) {
//Is the node in question a node call?
var asCall = nodeInQuestion as InvocationExpressionSyntax;
if (asCall != null) {
//Make sure we aren't on the right side of the call
if (asCall.Expression == leafNode)
return asCall;
MemberAccessExpressionSyntax memberAccessExpression = asCall.Expression as MemberAccessExpressionSyntax;
if (memberAccessExpression != null && memberAccessExpression.Name == leafNode)
return asCall;
GenericNameSyntax genericNameSyntax = asCall.Expression as GenericNameSyntax;
if (genericNameSyntax != null && genericNameSyntax.Identifier == leafToken)
return asCall;
return null;
}
var asProp = nodeInQuestion as MemberAccessExpressionSyntax;
if (asProp != null)
{
//Make sure we aren't on the left side of the dot
if (asProp.Name == lastNode && asProp.Name is IdentifierNameSyntax)
{
return asProp;
}
}
var asCtorCall = nodeInQuestion as ObjectCreationExpressionSyntax;
if (asCtorCall != null)
{
return asCtorCall;
}
var declNode = nodeInQuestion as MemberDeclarationSyntax;
if (declNode != null)
{
if (lastNode == leafNode) { return declNode; }
return null;
}
#warning What exactly is this doing?
#if false
NamedAssignmentNode na = nodeInQuestion as NamedAssignmentNode;
if (na != null)
{
if (na.Identifier == asName)
{
return na;
}
}
#endif
//Is our parent a binary operator?
var asBinaryOperator = nodeInQuestion as BinaryExpressionSyntax;
if (asBinaryOperator != null) {
//Make sure we are always on the rightmost of any binary operator
if (asBinaryOperator.Right != lastNonBinaryOperatorNode)
return null;
} else
lastNonBinaryOperatorNode = nodeInQuestion;
//Climb higher up our ancestry for the next iteration
lastNode = nodeInQuestion;
//.........这里部分代码省略.........