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<main>
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<h2 id="性能对比循环-vs-迭代器"><a class="header" href="#性能对比循环-vs-迭代器">性能对比：循环 VS 迭代器</a></h2>
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<blockquote>
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<p><a href="https://github.com/rust-lang/book/blob/main/src/ch13-04-performance.md">ch13-04-performance.md</a>
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<br>
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commit 009fffa4580ffb175f1b8470b5b12e4a63d670e4</p>
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</blockquote>
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<p>为了决定是否使用循环或迭代器，你需要了解哪个实现更快：使用显式 <code>for</code> 循环的 <code>search</code> 函数版本，还是使用迭代器的版本。</p>
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<p>我们进行了一个基准测试，将阿瑟·柯南·道尔的《福尔摩斯探案集》的全部内容加载到一个 <code>String</code> 中，并在内容中查找单词 “the”。以下是使用 <code>for</code> 循环版本和使用迭代器版本的 <code>search</code> 函数的基准测试结果：</p>
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<pre><code class="language-text">test bench_search_for ... bench: 19,620,300 ns/iter (+/- 915,700)
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test bench_search_iter ... bench: 19,234,900 ns/iter (+/- 657,200)
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</code></pre>
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<p>结果迭代器版本还要稍微快一点！这里我们不会解释性能测试的代码，我们的目的并不是为了证明它们是完全等同的，而是得出一个怎样比较这两种实现方式性能的基本思路。</p>
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<p>对于一个更全面的性能测试，你应该使用不同大小的文本作为 <code>contents</code>，不同的单词以及长度各异的单词作为 <code>query</code>，以及各种其他变化进行检查。关键在于：迭代器，作为一个高级的抽象，被编译成了与手写的底层代码大体一致性能的代码。迭代器是 Rust 的 <strong>零成本抽象</strong>（<em>zero-cost abstractions</em>）之一，它意味着抽象并不会引入额外的运行时开销，它与本贾尼·斯特劳斯特卢普（C++ 的设计和实现者）在 “Foundations of C++”（2012）中所定义的 <strong>零开销</strong>（<em>zero-overhead</em>）如出一辙：</p>
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<blockquote>
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<p>In general, C++ implementations obey the zero-overhead principle: What you don't use, you don't pay for. And further: What you do use, you couldn't hand code any better.</p>
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<ul>
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<li>Bjarne Stroustrup "Foundations of C++"</li>
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</ul>
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<p>从整体来说，C++ 的实现遵循了零开销原则：你不需要的，无需为它买单。更有甚者的是：你需要的时候，也无法通过手写代码做得更好。</p>
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<ul>
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<li>本贾尼·斯特劳斯特卢普 "Foundations of C++"</li>
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</ul>
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</blockquote>
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<p>作为另一个例子，以下代码取自一个音频解码器。解码算法使用线性预测数学运算（linear prediction mathematical operation）来根据之前样本的线性函数预测将来的值。这些代码使用迭代器链对作用域中的三个变量进行某种数学计算：一个叫 <code>buffer</code> 的数据 slice、一个有 12 个元素的数组 <code>coefficients</code>、和一个代表位数据位移量的 <code>qlp_shift</code>。我们在这个例子中声明了这些变量，但没有为它们赋值；虽然这些代码在其上下文之外没有太多意义，不过仍是一个简明的现实例子，来展示 Rust 如何将高级概念转换为底层代码。</p>
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<pre><code class="language-rust ignore">let buffer: &mut [i32];
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let coefficients: [i64; 12];
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let qlp_shift: i16;
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for i in 12..buffer.len() {
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let prediction = coefficients.iter()
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.zip(&buffer[i - 12..i])
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.map(|(&c, &s)| c * s as i64)
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.sum::<i64>() >> qlp_shift;
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let delta = buffer[i];
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buffer[i] = prediction as i32 + delta;
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}</code></pre>
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<p>为了计算 <code>prediction</code> 的值，这段代码遍历了 <code>coefficients</code> 中的 12 个值，使用 <code>zip</code> 方法将系数与 <code>buffer</code> 的前 12 个值组合在一起。接着将每一对值相乘，再将所有结果相加，然后将总和右移 <code>qlp_shift</code> 位。</p>
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<p>像音频解码器这样的程序通常最看重计算的性能。这里，我们创建了一个迭代器，使用了两个适配器，接着消费了其值。那么这段 Rust 代码将会被编译为什么样的汇编代码呢？好吧，在编写本书的这个时候，它被编译成与手写的相同的汇编代码。遍历 <code>coefficients</code> 的值完全用不到循环：Rust 知道这里会迭代 12 次，所以它“展开”（unroll）了循环。展开是一种将循环迭代转换为重复代码，并移除循环控制代码开销的代码优化技术。</p>
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<p>所有的系数都被储存在了寄存器中，这意味着访问它们非常快。这里也没有运行时数组访问边界检查。所有这些 Rust 能够提供的优化使得结果代码极为高效。现在你知道了这些，请放心大胆的使用迭代器和闭包吧！它们使得代码看起来更高级，但并不为此引入运行时性能损失。</p>
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<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
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<p>闭包和迭代器是 Rust 受函数式编程语言观念所启发的功能。它们对 Rust 以高性能来明确的表达高级概念的能力有很大贡献。闭包和迭代器的实现达到了不影响运行时性能的程度。这正是 Rust 致力于提供零成本抽象的目标的一部分。</p>
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<p>现在我们改进了 I/O 项目的（代码）表现力，那么让我们来看看 <code>cargo</code> 的更多功能，这些功能将帮助我们将项目分享给全世界。</p>
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