blocks|key|2716568|text|这是我能想到的最好的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2716569|func+roundedInt<T:+FloatingPoint>(_+f:+T)+->+Int+{
++++return+Int(Float80("\(f.rounded())")!)
}|code-block|syntax|javascript|2716570|基本上，这是一种“欺骗”。据我所知，标准库中符合FloatingPoint的所有类型都是Float、Double和Float80。它们的字符串表示形式是相同的。这就是为什么我将字符串表示转换为Float80的原因。然后，我将这个数字转换为一个Int。|offset|length|style|CODE|2716571|我说这是一种欺骗，因为它实际上不适用于大量的数字。当您将很大的数字放入此函数中时，会发生致命错误。经过一些试验和错误之后，我发现在64位处理器上可以传递给这个函数的最大数字是9223372036854775295，比Int.max少512个。|2716572|此外，如果您在那里通过9223372036854775295，它将返回9223372036854770000，而不是原来的数字。这很可能是因为浮点数不准确。|2716573|entityMap^0|0|0|O|D|18|5|1E|6|1L|7|2O|7|3D|3|0|30|7|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|R|$]]

This is the best I can come up with:

<pre><code>func roundedInt&lt;T: FloatingPoint&gt;(_ f: T) -&gt; Int {
 return Int(Float80("\(f.rounded())")!)
}
</code></pre>

Basically, this is kind of a "cheat". As far as I know, all the types in the standard library that conform to <code>FloatingPoint</code> are <code>Float</code>, <code>Double</code> and <code>Float80</code>. Their string representations are in the same format. This is why I converted the string representation to a <code>Float80</code>. I then converted this number to an <code>Int</code>.

I say this is a cheat because it does not really work for large numbers. When you put really large numbers into this function, a fatal error occurs. After some trial and error, I found out that the largest number you can pass to this function is 9223372036854775295 on a 64-bit processor, which is 512 less than <code>Int.max</code>.

Also, if you pass 9223372036854775295 in there, it will return 9223372036854770000 instead of the original number. This is most likely because floating points are inaccurate.

blocks|key|2716596|text|简单的C%2B%2B模板只不过是一种宏、类型少的源代码替换机制.调用代码被应用模板替换，编译器将检查生成的代码是否有意义。您是对的，roundedInt<T>无界函数在C%2B%2B域应该可以很好地工作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2716597|相反，Swift泛型是按设计设计的类型，这意味着泛型代码必须是独立的，独立于给定呼叫站点的任何细节。在您的示例中，FloatingPoint协议是引导编译的类型(而不是调用代码使用的实际T类型)。|2716598|(顺便说一下，Java/C#泛型也类似Swift风格。)|2716599|对于实际问题，只需为roundedInt函数提供两个重载：|2716600|func+roundedInt(_+f:+Float)++->+Int+{+...+}
func+roundedInt(_+f:+Double)+->+Int+{+...+}|code-block|syntax|javascript|2716601|这应该涵盖大多数用例。当然，假设您只是在编写小助手函数.而不是完整的库/框架！|2716602|否则，尝试@Sweeper基于字符串的解决方案。但请记住，除了潜在的精度损失，他正确地指出，也有一些恶劣的性能问题潜伏在那里。|2716603|entityMap^0|1Q|D|0|1L|D|2L|1|0|0|A|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|L|3|M|5|N|7|18|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1A|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|V|$]]

Simple C++ templates are nothing more than a macro-ish, type less, source code replacement mechanism. The calling code is replaced with the applied template and the compiler checks if the resulting code makes any sense. You are right, a <code>roundedInt&lt;T&gt;</code> unbounded function should work fine in C++ land. 

Swift generics instead are typesafe by design, meaning the generic code must be sound on its own, independently of any particulars of a given call site. In your example, the <code>FloatingPoint</code> protocol is the type guiding the compilation (and not the actual <code>T</code> type used by the calling code). 

(By the way, Java/C# generics also resembles Swift style as well.)

<hr>

Regarding your actual problem, you could simply provide two overloads for the <code>roundedInt</code> function: 

<pre><code>func roundedInt(_ f: Float) -&gt; Int { ... }
func roundedInt(_ f: Double) -&gt; Int { ... }
</code></pre>

which should cover most use cases. Of course, assuming you are only writing small helper functions with this... and not full blown libraries/frameworks!

Otherwise, try @Sweeper string-based solution. But please keep in mind that besides the potential loss of precision he correctly noted, there are also some nasty performance problems lurking in there as well.

blocks|key|2716612|text|使用BinaryFloatingPoint，正如您所期望的那样，在Int可以表示的范围内工作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2716613|func+roundedInt<T:+BinaryFloatingPoint>(_+f:+T)+->+Int+{
+++return+Int(f.rounded())
}

let+x+=+roundedInt(Float(42.1))+//+42|code-block|syntax|javascript|2716614|entityMap^0|2|J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Using <code>BinaryFloatingPoint</code> this works as you expect, within the bounds of what Int can represent.

<pre><code>func roundedInt&lt;T: BinaryFloatingPoint&gt;(_ f: T) -&gt; Int {
 return Int(f.rounded())
}

let x = roundedInt(Float(42.1)) // 42
</code></pre>

Consider the following (simplified from actual code):

<pre><code>func roundedInt&lt;T: FloatingPoint&gt;(_ f: T) -&gt; Int {
 return Int(f.rounded())
}
</code></pre>

This fails to compile with the following error:

<blockquote>
 Cannot invoke initializer for type 'Int' with an argument list of type
 '(T)' Overloads for 'Int' exist with these partially matching
 parameter lists: (Int64), (Word), (UInt8), (Int8), (UInt16), (Int16),
 (UInt32), (Int32), (UInt64), (UInt), (Int), (Float), (Double),
 (Float80), (String, radix: Int), (CGFloat), (NSNumber)
</blockquote>

I take this to mean that FloatingPoint can't be matched to any of the Int overloads.

Is there a simple way to make this work? Or is there a limitation to Swift generics (e.g., compared to C++ templates) that precludes this?

Swift: Casting a generic, FloatingPoint value to Int

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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考虑以下几点(从实际代码中简化)：func roundedInt<T: FloatingPoint>(_ f: T) -> Int { return Int(f.rounded())}这在编译过程中出现了以下错误：不能使用“Int”类型的参数列表调用“Int”类型的初始化器，因为“Int”的重载包含这些部分匹配的...

问Swift:将泛型的FloatingPoint值转换为Int
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问Swift:将泛型的FloatingPoint值转换为IntEN