为什么不能用VAR返回类型声明一个函数？ [英] Why is it not possible to declare a function with VAR return type?

问题描述

在C＃中，我们有 var 数据类型，但我们不能将其用作函数返回类型。

为什么不可能？ / p>

In C#, we have var data type but we can't use it as functions return type.
Why this is not possible?

public var myFunction()
{
    var = some operations
}

推荐答案

我认为这部分是由于编译器的设计。 Eric Lippert 博客了解为什么字段不能使用隐式类型，我怀疑一些相同的参数保存方法。

I believe it's partly due to the design of the compiler. Eric Lippert blogged about why fields can't use implicit typing, and I suspect some of the same arguments hold for methods.

但是你可能很容易结束歧义中。例如：

But you could easily end up with ambiguity anyway. For example:

var Method1(bool callMethod2)
{
    return callMethod2 ? Method2() : null;
}

var Method2()
{
    return Method1(false);
}

这里应该是什么类型？

What should the type be here?

一个更简单的例子：

var Method1(bool throwException)
{
    if (!throwException)
    {
        return Method1(true);
    }
    throw new Exception("Bang!");
}

诚然，这种歧义可以简单地被禁止，但我怀疑设计团队认为设计和实施的复杂性增加不值得。不要忘记它们使用有限的资源运行 - 给方法和 async / await 之间的 var ，我会选择后者在心跳。 （诚​​然，还有其他的功能，我会选择，而不是动态，但这是一个不同的事情...）

Admittedly this sort of ambiguity could simply be disallowed, but I suspect that the design team felt that the added complexity of both design and implementation wasn't worth the benefit. Don't forget that they're running with limited resources - given a choice between var for methods and async/await, I'd pick the latter in a heartbeat. (Admittedly there are other features I'd have picked instead of dynamic, but that's a different matter...)

请注意，返回类型推断是对lambda表达式执行，所以它的想法不是疯了。例如：

Note that return type inference is performed for lambda expressions, so the very idea of it isn't crazy. For example:

IEnumerable<string> x = new[] { "x", "y", "z" };

var result = x.Select(s => { return s.Length; }); // Long form

编译器在执行重载分辨率时推断出完整类型的lambda表达式选择，将其转换为 Func< string，int> 。将同样的想法应用于方法并不难想象 - 只是复杂。

There the compiler infers the complete type of the lambda expression when it performs overload resolution on Select, converting it to a Func<string, int>. It's not inconceivable to apply the same ideas to methods - just complicated.

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