[英]Converting std::array to std::vector
In the code below, the size of the function (foo) argument ( std::vector
) can be anything to make the function a generic one. 在下面的代码中,函数(foo)参数(
std::vector
)的大小可以是使函数成为通用函数的任何东西。 However, sometimes the size container is known so std::array can be used. 但是,有时大小容器是已知的,因此可以使用std :: array。 The problem is to convert the
std::array
to std::vector
. 问题是将
std::array
转换为std::vector
。 What is the best way to solve this problem? 解决这个问题的最佳方法是什么? Is it better to just use
std::vector
always in this case? 在这种情况下,总是使用
std::vector
总是更好吗?
#include <iostream>
#include <array>
#include <vector>
using namespace std;
// generic function: size of the container can be anything
void foo (vector<int>& vec)
{
// do something
}
int main()
{
array<int,3> arr; // size is known. why use std::vector?
foo (arr); // cannot convert std::array to std::vector
return 0;
}
Given that you pass in an array
, foo
does not seem to resize the container it gets passed in (however, it does seem to modify the elements since vec
is passed non-const). 鉴于你传入一个
array
, foo
似乎没有调整它传入的容器的大小(但是,它似乎修改了元素,因为vec
是非const传递的)。 It therefore does not need to know anything about the underlying container other than how to access elements. 因此,除了如何访问元素之外,它不需要知道任何有关底层容器的信息。
Then, you can pass a pair of iterators (and make the iterator type a template argument), as many STL algorithms do. 然后,您可以传递一对迭代器 (并使迭代器类型成为模板参数),就像许多STL算法一样。
The function accepts a vector by reference. 该函数通过引用接受向量。 So there is no any sense to pass an object of type std::array to the function.
因此将std :: array类型的对象传递给函数没有任何意义。 It could have a sense if the parameter of the function would be defined as const reference to a vector.
如果函数的参数被定义为向量的const引用,则可能有意义。
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