1 write to _languages
Microsoft.DotNet.SwaggerGenerator.CodeGenerator (1)
Languages\Language.cs (1)
56
_languages
= new Dictionary<string, Language>(StringComparer.OrdinalIgnoreCase)
1 reference to _languages
Microsoft.DotNet.SwaggerGenerator.CodeGenerator (1)
Languages\Language.cs (1)
67
_languages
.TryGetValue(name, out Language value);