C# Reactive Extensions (rx) FirstOrDefault 枚举整个集合

Question

似乎 FirstOrDefault 的预期行为是在找到与谓词匹配的项目后完成，而 concat 的预期行为是懒惰地评估。但是，即使谓词与第一项匹配，以下示例也会枚举整个集合。

（感谢更友好的代码 Shlomo）

void Main()
{
    var entities = Observable.Defer(() => GetObservable().Concat());
    Entity result = null;
    var first = entities.FirstOrDefaultAsync(i => i.RowId == 1).Subscribe(i => result = i);
    result.Dump();
    buildCalled.Dump();
}

// Define other methods and classes here

public IEnumerable<IObservable<Entity>> GetObservable()
{
    var rows = new List<EntityTableRow>
    {
        new EntityTableRow { Id = 1, StringVal = "One"},
        new EntityTableRow { Id = 2, StringVal = "Two"},
    };
    return rows.Select(i => Observable.Return(BuildEntity(i)));
}

public int buildCalled = 0;
public Entity BuildEntity(EntityTableRow entityRow)
{
    buildCalled++;
    return new Entity { RowId = entityRow.Id, StringVal = entityRow.StringVal };
}

public class Entity
{
    public int RowId { get; set; }
    public string StringVal { get; set; }
}

public class EntityTableRow
{
    public int Id { get; set; }
    public string StringVal { get; set; }
}

这是预期的行为吗？有没有办法将对象（特别是本例中的建筑物）的枚举推迟到真正需要时？

Answer 1

以下是与您所拥有的等价的 Linqpad 友好代码：

void Main()
{
    var entities = Observable.Defer(() => GetObservable().Concat());
    Entity result = null;
    var first = entities.FirstOrDefaultAsync(i => i.RowId == 1).Subscribe(i => result = i);
    result.Dump();
    buildCalled.Dump();
}

// Define other methods and classes here

public IEnumerable<IObservable<Entity>> GetObservable()
{
    var rows = new List<EntityTableRow>
    {
        new EntityTableRow { Id = 1, StringVal = "One"},
        new EntityTableRow { Id = 2, StringVal = "Two"},
    };
    return rows.Select(i => Observable.Return(BuildEntity(i)));
}

public int buildCalled = 0;
public Entity BuildEntity(EntityTableRow entityRow)
{
    buildCalled++;
    return new Entity { RowId = entityRow.Id, StringVal = entityRow.StringVal };
}

public class Entity
{
    public int RowId { get; set; }
    public string StringVal { get; set; }
}

public class EntityTableRow
{
    public int Id { get; set; }
    public string StringVal { get; set; }
}

如果您将GetObservable 更改为以下内容，您将获得所需的结果：

public IObservable<IObservable<Entity>> GetObservable()
{
    var rows = new List<EntityTableRow>
    {
        new EntityTableRow { Id = 1, StringVal = "One"},
        new EntityTableRow { Id = 2, StringVal = "Two"},
    };
    return rows.ToObservable().Select(i => Observable.Return(BuildEntity(i)));
}

似乎Concat<TSource>(IEnumerable<IObservable<TSource>>) 的实现渴望评估可枚举，而Concat<TSource>(IObservable<IObservable<TSource>>) 和ToObservable<TSource>(IEnumerable<TSource>) 的实现适当地保持惰性。我不能说我知道为什么。