object.ReferenceEquals 或 == 运算符？

发布于 2024-11-30 12:45:22 字数 603 浏览 1 评论 0原文

为什么 ThrowIfNull 实现为：

    static void ThrowIfNull<T>(this T argument, string name) where T : class
    {
        if (argument == null)
        {
            throw new ArgumentNullException(name);
        }
    }

重写为：

    static void ThrowIfNull<T>(this T argument, string name) where T : class
    {
        if (object.ReferenceEquals(argument, null))
        {
            throw new ArgumentNullException(name);
        }
    }

优点：它有助于避免令人困惑的 Equals 重载，并且可能使代码更加清晰。

这样做有什么缺点吗？应该有一些。

原文

Why is ThrowIfNull implemented as:

    static void ThrowIfNull<T>(this T argument, string name) where T : class
    {
        if (argument == null)
        {
            throw new ArgumentNullException(name);
        }
    }

Wouldn't it be better rewritten as:

    static void ThrowIfNull<T>(this T argument, string name) where T : class
    {
        if (object.ReferenceEquals(argument, null))
        {
            throw new ArgumentNullException(name);
        }
    }

Pros: it helps avoiding confusing Equals overloads and probably makes the code more clear.

Any cons to that? There should be some.

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时光暖心i 2024-12-07 12:45:22

两者没有区别。您将覆盖 Equals（在任一实现中都没有调用）与重载 ==（其中在这两个代码段中都无关紧要，因为重载是在编译时执行的，并且编译器对 T 的了解不够，无法使用任何特定的重载）。

只是为了说明我的意思：

static void ThrowIfFoo<T>(this T argument, string name) where T : class
{
    if (argument == "foo")
    {
        throw new Exception("You passed in foo!");
    }
}

测试使用：

"foo".ThrowIfFoo(); // Throws

string x = "f";
x += "oo"; // Ensure it's actually a different reference

x.ThrowIfFoo(); // Doesn't throw

ThrowIfFoo 不知道 T 将是一个字符串 - 因为这取决于调用代码- 并且重载决策仅在编译 ThrowIfFoo 时执行。因此，它使用运算符 ==(object, object) 而不是 ==(string, string)。

换句话说，它是这样的：

object foo1 = "foo";

string tmp = "f";
object foo2 = tmp + "oo";

Console.WriteLine(foo1.Equals(foo2)); // Prints True
Console.WriteLine(foo1 == foo2); // Prints false
Console.WriteLine((string) foo1 == (string) foo2); // Prints True

在最后一行，编译器知道它可以使用 == 的重载，因为两个操作数都具有 string 的编译时类型。

There's no difference between the two. You're confusing overriding Equals (which isn't called in either implementation) with overloading == (which won't be relevant in either snippet as overloading is performed at compile time, and the compiler doesn't know enough about T to use any specific overload).

Just to show what I mean:

static void ThrowIfFoo<T>(this T argument, string name) where T : class
{
    if (argument == "foo")
    {
        throw new Exception("You passed in foo!");
    }
}

Testing with:

"foo".ThrowIfFoo(); // Throws

string x = "f";
x += "oo"; // Ensure it's actually a different reference

x.ThrowIfFoo(); // Doesn't throw

ThrowIfFoo doesn't know that T will be a string - because that depends on the calling code - and the overload resolution is only performed when ThrowIfFoo is compiled. Therefore it's using the operator ==(object, object) rather than ==(string, string).

In other words, it's like this:

object foo1 = "foo";

string tmp = "f";
object foo2 = tmp + "oo";

Console.WriteLine(foo1.Equals(foo2)); // Prints True
Console.WriteLine(foo1 == foo2); // Prints false
Console.WriteLine((string) foo1 == (string) foo2); // Prints True

In the last line, the compiler knows it can use the overload of == because both operands have compile-time types of string.

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