4 references to Show
System.Windows.Forms.Design (2)
System\ComponentModel\Design\DesignerActionUI.cs (1)
390
if (e.ChangeType == DesignerActionUIStateChangeType.
Show
)
System\ComponentModel\Design\DesignerActionUIService.cs (1)
54
new DesignerActionUIStateChangeEventArgs(component, DesignerActionUIStateChangeType.
Show
));
System.Windows.Forms.Design.Tests (2)
System\ComponentModel\Design\DesignerActionUIStateChangeEventArgsTests.cs (2)
10
yield return new object[] { null, DesignerActionUIStateChangeType.
Show
- 1 };
11
yield return new object[] { new(), DesignerActionUIStateChangeType.
Show
};