所有容器都具有.size()函数吗? [英] Do all containers have a .size() function?

问题描述

对于家庭作业，我必须创建一个可以在任何容器上执行的模板化标准差函数.这就是我所拥有的:

For a homework assignment I have to create a templatized standard deviation function that can be performed on any container. Here's what I have:

template <typename Container>
double findMean(Container c, int count){
    double sum = 0;
    for (auto&& e : c){
        sum += e;
    }
    sum /= count;
    return sum;
}

template <typename Container>
double findStDev(Container c){
    double mean = findMean(c, c.size());
    std::cout << mean << std::endl;
    for (auto&& e : c){
        e -= mean;
        e *= e;
    }
    mean = sqrt(findMean(c, c.size()));
    return mean;
}

第一次找到均值要除以容器的全部大小(n)，但是第二次找到标准差时，我需要除以size-1(n-1)).

The first time I find the mean I want to divide by the full size of the container (n), but when I find it the second time for the standard deviation, I need to divide by size-1 (n-1).

.size()函数是否可用于所有c ++容器?

Is the .size() function available for all c++ containers?

推荐答案

几乎.根据表96-N3797中的容器要求，标准库中的所有容器都必须提供成员函数 size .它应具有恒定的执行时间，并返回容器 a 的 distance(a.begin()，a.end())值.

Almost. By table 96 - container requirements in N3797, all containers in the standard library must provide a member function size. It shall have constant execution time and return the value of distance(a.begin(),a.end()) for a container a.

但是，稍后将提到一个(只有一个)例外:

However, there is one (and only one) exception mentioned later on:

forward_list满足容器的所有要求(表96)，但size()成员除外功能未提供.

A forward_list satisfies all of the requirements of a container (Table 96), except that the size() member function is not provided.

(N3797 23.3.4.1第2条)

(N3797 23.3.4.1 Clause 2)

这意味着 std :: forward_list 实际上是一个标准容器，没有 没有成员函数 size .

That means that std::forward_list is indeed a standard container that does not have a member function size.

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