SyntaxNode.DescendantTrivia Method
Definition
Important
Some information relates to prerelease product that may be substantially modified before it’s released. Microsoft makes no warranties, express or implied, with respect to the information provided here.
Overloads
DescendantTrivia(TextSpan, Func<SyntaxNode,Boolean>, Boolean) |
Get a list of all the trivia associated with the descendant nodes and tokens. |
DescendantTrivia(Func<SyntaxNode,Boolean>, Boolean) |
Get a list of all the trivia associated with the descendant nodes and tokens. |
DescendantTrivia(TextSpan, Func<SyntaxNode,Boolean>, Boolean)
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
Get a list of all the trivia associated with the descendant nodes and tokens.
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxTrivia> DescendantTrivia (Microsoft.CodeAnalysis.Text.TextSpan span, Func<Microsoft.CodeAnalysis.SyntaxNode,bool> descendIntoChildren = default, bool descendIntoTrivia = false);
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxTrivia> DescendantTrivia (Microsoft.CodeAnalysis.Text.TextSpan span, Func<Microsoft.CodeAnalysis.SyntaxNode,bool>? descendIntoChildren = default, bool descendIntoTrivia = false);
member this.DescendantTrivia : Microsoft.CodeAnalysis.Text.TextSpan * Func<Microsoft.CodeAnalysis.SyntaxNode, bool> * bool -> seq<Microsoft.CodeAnalysis.SyntaxTrivia>
Public Function DescendantTrivia (span As TextSpan, Optional descendIntoChildren As Func(Of SyntaxNode, Boolean) = Nothing, Optional descendIntoTrivia As Boolean = false) As IEnumerable(Of SyntaxTrivia)
Parameters
- span
- TextSpan
- descendIntoChildren
- Func<SyntaxNode,Boolean>
- descendIntoTrivia
- Boolean
Returns
Applies to
DescendantTrivia(Func<SyntaxNode,Boolean>, Boolean)
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
Get a list of all the trivia associated with the descendant nodes and tokens.
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxTrivia> DescendantTrivia (Func<Microsoft.CodeAnalysis.SyntaxNode,bool> descendIntoChildren = default, bool descendIntoTrivia = false);
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxTrivia> DescendantTrivia (Func<Microsoft.CodeAnalysis.SyntaxNode,bool>? descendIntoChildren = default, bool descendIntoTrivia = false);
member this.DescendantTrivia : Func<Microsoft.CodeAnalysis.SyntaxNode, bool> * bool -> seq<Microsoft.CodeAnalysis.SyntaxTrivia>
Public Function DescendantTrivia (Optional descendIntoChildren As Func(Of SyntaxNode, Boolean) = Nothing, Optional descendIntoTrivia As Boolean = false) As IEnumerable(Of SyntaxTrivia)
Parameters
- descendIntoChildren
- Func<SyntaxNode,Boolean>
- descendIntoTrivia
- Boolean
Returns
Applies to
Váš názor
https://aka.ms/ContentUserFeedback.
Připravujeme: V průběhu roku 2024 budeme postupně vyřazovat problémy z GitHub coby mechanismus zpětné vazby pro obsah a nahrazovat ho novým systémem zpětné vazby. Další informace naleznete v tématu:Odeslat a zobrazit názory pro