std::transform 使用 C++0x lambda 表达式

std::transform using C++0x lambda expression

提问by Steve Townsend

How is this done in C++0x?

这是如何在 C++0x 中完成的？

std::vector<double> myv1;
std::transform(myv1.begin(), myv1.end(), myv1.begin(),
               std::bind1st(std::multiplies<double>(),3));

Original question and solution is here.

原始问题和解决方案在这里。

回答by Edward Strange

std::transform(myv1.begin(), myv1.end(), myv1.begin(), 
   [](double d) -> double { return d * 3; });

回答by Steve Jessop

The main original motivation for using that functional style for these cases in C++ was, "aaagh! iterator loops!", and C++0x removes that motivation with the range-based for statement. I know that part of the point of the question was to find out the lambda syntax, but I think the answer to the question "How is this done in C++0x?" is:

在 C++ 中对这些情况使用这种函数式风格的主要原始动机是，“aaagh！迭代器循环！”，而 C++0x 使用基于范围的 for 语句消除了这种动机。我知道问题的一部分是找出 lambda 语法，但我认为这个问题的答案“这是如何在 C++0x 中完成的？” 是：

for(double &a : myv1) { a *= 3; }

There's no actual function object there, but if it helps you could pretend that { a *= 3; }is a highly abbreviated lambda. For usability it amounts to the same thing either way, although the draft standard defines range-based for in terms of an equivalent forloop.

那里没有实际的函数对象，但如果它有帮助，您可以假装这{ a *= 3; }是一个高度缩写的 lambda。就可用性而言，无论哪种方式都是相同的，尽管草案标准根据等效for循环定义了基于范围的 for 。

回答by GManNickG

Just do as Dario says:

就像达里奥说的那样：

for_each(begin(myv1), end(myv1), [](double& a) { a *= 3; });

for_eachis allowed to modify elements, saying it cannot is a myth.

for_each允许修改元素，说它不能是一个神话。

回答by Dario

Using a mutable approach, we can use for_eachto directly update the sequence elements through references.

使用可变方法，我们可以for_each通过引用直接更新序列元素。

for_each(begin(myv1), end(myv1), [](double& a) { a *= 3; });

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是否for_eachfor_each真的允许修改元素一直存在一些争论，因为它被称为“非变异”算法。

What that means is for_eachisn't allowed to alter the sequenceit operates on (which refers to changes of the sequence structure- i.e. invalidating iterators). This doesn't mean we cannot modify the non-const elementsof the vector as usual - the structure itself is left untouched by these operations.

这意味着for_each不允许改变它操作的序列（这是指序列结构的变化——即使迭代器无效）。这并不意味着我们不能像往常一样修改向量的非常量元素- 这些操作不会影响结构本身。

回答by John Dibling

Like this:

像这样：

vector<double> myv1;
transform(myv1.begin(), myv1.end(), myv1.begin(), [](double v)
{
    return v*3.0;
});

回答by Elligno

I'm using VS2012 which support the C++11 bind adaptor. To bind the first element of the binary function (as bind1st use to do) you need to add an _1 (placeholder argument). Need to include the functional for bind.

我正在使用支持 C++11 绑定适配器的 VS2012。要绑定二元函数的第一个元素（如 bind1st 使用的那样），您需要添加一个 _1（占位符参数）。需要包含用于绑定的函数。

using namespace std::placeholders;
std::transform( myv1.begin(), myv1.end(), myv1.begin(),
                 std::bind( std::multiplies<double>(),3,_1));