Question

我有以下类型推断“失败”的情况（至少失败了我希望它做了什么）。基本上，我有一个方法接受泛型类型的数组。我需要将该数组类型化为匿名对象，但类型推断无法执行此操作。

    private void foo<T>(IEnumerable<T> items, Func<T, Object>[] propertySelector) { }

    public void Main()
    {
        var peeps = new[]
        {
            new {FirstName = "Taco", LastName = "King"},
            new {FirstName = "Papa", LastName = "Georgio"}
        };

        foo(peeps, new[]
        {
            an => an.FirstName, //Error cannot infer type of "an"
            an => an.LastName   //Error cannot infer type of "an"
        });
    }

我认为原因是因为数组类型是从其内容而不是其上下文推断出来的。这似乎使得在这种情况下无法使用匿名类型。

有什么想法绕过这个？

Answer 1

在给出的示例中，您可以将propertySelector更改为params参数，然后单独传递每个函数而不是数组。如果由于某种原因无法使用params，那么这样的辅助函数将起作用：

    /// <summary>
    /// Allows the use of type inference to get selector functions for the type of an enumerable.
    /// </summary>
    /// <typeparam name="T">The type of the enumerable.</typeparam>
    /// <param name="enumerable">The enumerable.</param>
    /// <param name="selectors">A set of selectors to return.</param>
    /// <returns>The selectors passed in.</returns>
    public static Func<T, Object>[] GetSelectors<T>(
        IEnumerable<T> enumerable,
        params Func<T, Object>[] selectors)
    {
        return selectors;
    }

所以你的例子会变成：

private void foo<T>(IEnumerable<T> items, Func<T, Object>[] propertySelector) { }

public void Main()
{
    var peeps = new[]
    {
        new {FirstName = "Taco", LastName = "King"},
        new {FirstName = "Papa", LastName = "Georgio"}
    };

    foo(peeps, GetSelectors(peeps, an => an.FirstName, an => an.LastName));
}

类型推断失败

1 个答案: