无法从D中的std.range切片Take！R吗？

问题描述

我正在尝试使用slice运算符从std.range获取take函数的返回值的分片。我的代码：

I'm trying to use the slice operator to obtain a slice of the return value of the take function from std.range. My code:

auto tempChunk = ['a', 'b', 'c', 'd'];
auto a = tempChunk.take(3);
writeln(a[0..2]);

在这种情况下，作为Take！R只是char []的别名，我希望这个编译。但是，编译器告诉我 Take！（char []）无法用[] 切片。再举一个例子：

As Take!R in this case is just an alias for char[], I'd expect this to compile. However, the compiler tells me that Take!(char[]) cannot be sliced with []. Taking another example:

int[] arr1 = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]; 
auto s = arr.take(5);
writeln(s[0..4]);

这将编译并运行，没有问题，打印[1、2、3、4、5] 。关于第一个示例为什么不起作用而第二个示例为什么不起作用，我完全感到困惑。

This will compile and run without a problem, printing [1, 2, 3, 4, 5]. I am completely confused at this point as to why the first example won't work, while the second does.

推荐答案

take 模板使用 hasSlicing 确定是否可以返回输入的片段而不是Take！R结构。检查实际的返回类型使它更加清晰：

take template uses hasSlicing to determine if slice of input can be returned instead of Take!R struct. Checking actual return types makes it a bit more clear:

import std.range, std.stdio;

void main()
{
    auto chararr = ['a', 'b', 'c', 'd'];
    auto a = chararr.take(3);
    writeln( typeid(typeof(a)) );

    auto intarr = [ 1, 2, 3, 4 ];  
    auto b = intarr.take(3);
    writeln( typeid(typeof(b)) );
}

// Output:
// std.range.Take!(char[]).Take
// int[]

已进行切片明确指示为所有窄字符串返回 false -这些元素可能不代表单个字符，而是一个代码点（基于char和wchar的字符）。

hasSlicing is explicitly instructed to return false for all "narrow strings" - those, which element may not represent single character, but a code point (char and wchar based ones).

现在，这是我的猜测开始的地方，但我想这样做是为了防止意外生成格式错误的UTF-8和使用切片的Co字符串。如果您在char []中没有任何实际需要，最好使用dchar []。

Now, here is where my speculations start but I suppose it is done to prevent accidental creating of malformed UTF-8 & Co strings using slicing. Better to use dchar[] if you do not have any real need in char[].

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