英文:
How to construct generic type from symbol info in roslyn
问题
INamedTypeSymbol - System.Action<MyAwsomeApp.Foo>.
英文:
I have INamedTypeSymbol, lets say MyAwsomeApp.Foo. I would like to wrap it in Action<>. How I can achieve this?
Expected output is INamedTypeSymbol - System.Action<MyAwseomeApp.Foo>.
答案1
得分: 1
你可以使用 INamedType.Construct 方法来实现这个，例如：
var actionOfT = compilation.GetTypeByMetadataName(typeof(Action<>).FullName);
var foo = compilation.GetTypeByMetadataName("MyAwseomeApp.Foo");
var actionOfFoo = actionOfT.Construct(foo);
英文:
You can use the INamedType.Construct method for that, e.g.
var actionOfT = compilation.GetTypeByMetadataName(typeof(Action<>).FullName);
var foo = compilation.GetTypeByMetadataName("MyAwseomeApp.Foo");
var actionOfFoo = actionOfT.Construct(foo);
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