转换列表＆lt; Guid＆gt;列出＆lt; GUID？ ＆GT;

Question

转换List＆lt;的最佳做法是什么？ Guid＆gt;列出＆lt; GUID？ ＆GT;

以下内容无法编译：

public List<Guid?> foo()
{
    List<Guid> guids = getGuidsList();
    return guids; 
}

Answer 1

问题似乎改变了几次，所以我会双向展示转换：

将List<Guid?>转换为List<Guid>：

var guids = nullableGuids.OfType<Guid>().ToList();
// note that OfType() implicitly filters out the null values,
// a Cast() would throw a NullReferenceException if there are any null values

将List<Guid>转换为List<Guid?>：

var nullableGuids = guids.Cast<Guid?>().ToList();

Answer 2

public List<Guid> foo()
{
    return  foo.Where(x=>x != null).Cast<Guid>().ToList();
}

Answer 3

像这样的东西

return guids.Select(e => new Guid?(e)).ToList();

Answer 4

略有不同的方法：

public List<Guid> foo()
{
    return foo.Where(g => g.HasValue).Select(g => g.Value).ToList();
}

Answer 5

public List<Guid?> foo()
{
    List<Guid> source = getGuidsList();
    return  source.Select(x => new Guid?(x)).ToList();

}