是否可以将插值字符串作为参数传递给方法？ [英] Is it possible to pass interpolated strings as parameter to a method?

问题描述

我已经开始使用插值字符串（新的C＃功能6），它是非常有用和优雅。但根据我的需要我必须要通过字符串格式作为参数的方法。喜欢的东西下一：

I have started to use Interpolated Strings (new feature of C# 6) and it is really useful and gracefully. But according to my needs I have to pass format of string to a method as a parameter. Something like next:

MyMethod(string format)

在过去，我用它在未来的方式：

In the past, I used it in the next way:

MyMethod("AAA{0:00}")

现在我想这个代码：

MyMethod($"AAA{i:00}")

但是，这并不工作，因为 I 方法内创建，并超出范围在这种情况下。

But this doesn't work, because i is created inside of the method and is out of scope in this context.

是否有可能使用任何伎俩通过插值字符串作为参数的方法？

Is it possible to use any trick for passing interpolated strings as a parameter to a method?

推荐答案

您不能这样做，这不会是一个很好的主意 - 这意味着你使用的是局部变量从另一个方法结果
这会破坏这一功能的目的 - 有编译器可核实的字符串插值和有约束力的局部变量。

You cannot do that, and that would not be a very good idea either - it means you are using local variables from another method.
This would defeat the purpose of this feature - to have compiler verifiable string interpolation and binding to local variables.

C＃有几个很好的替代品。例如，使用一个函数功能：

C# has several good alternatives. For example, using a Func:

public void MyMethod(Func<int,string> formatNumber)
{
    int i = 3;
    var formatted = formatNumber(i);
}

使用为：

MyMethod(number => $"AAA{number:00}");

这是比你有什么今天更好 - 那就是你有格式化字符串（？S）和其。使用在不同的地方

This is better than what you have today - where you have the format string(s?) and its usage in different places.

如果你有超过一个变量这种方法可以扩展，但认为它们组合到一个类（或结构）更多 - 对 FUNC 看起来好多了，你的代码将更具可读性。这个类也可以覆盖的ToString（），这可能是对你有用。

If you have more than a single variable this approach can scale, but consider grouping them into a class (or a struct) - the func will look much better, and your code will be more readable. This class can also override .ToString(), which might be useful for you.

另一个简单的办法是通过只是格式为：的MyMethod（00）和 i.ToString（格式），但是，只有在工作您大概简单的例子。

Another simple option is to pass just the format: MyMethod("00") and i.ToString(format), but that only works in your presumably simplified example.

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