std :: experimental :: ostream_joiner和std :: pair [英] std::experimental::ostream_joiner and std::pair
问题描述
在c ++ 17/g ++ 7中,终于有了长时间错过的ostream_joiner.它可以正确输出到ostream,并使用中界定界符分隔集合元素.
In c++17/g++7, there's finally the long missed ostream_joiner. It enables proper output to ostreams, separating collection elements with infix delimiters.
#include <algorithm>
#include <experimental/iterator>
#include <iostream>
#include <iterator>
#include <vector>
#include <string>
using string = std::string;
#if 1
struct pair {
string first;
string second;
};
#else
using pair = std::pair<string,string>;
#endif
std::ostream& operator<<(std::ostream& lhs, const pair &p) {
return lhs << p.first << "=" << p.second;
}
int main()
{
std::vector<pair> pairs = {{"foo", "bar"}, {"baz", "42"}};
std::copy(std::begin(pairs),
std::end(pairs),
std::experimental::make_ostream_joiner(std::cout, ", "));
}
代码段成功编译并输出...
Whilst the code piece succesfully compiles and outputs ...
foo=bar, baz=42
...在代码段中将#if 1更改为#if 0使编译器抱怨缺少适当的移位运算符:
... changing the #if 1 to a #if 0 in the snippet makes the compiler complaining about missing the proper shift operator:
main.cpp:29:70: required from here
/usr/local/include/c++/7.2.0/experimental/iterator:88:10: error: no match for
'operator<<' (operand types are
'std::experimental::fundamentals_v2::ostream_joiner<const char*, char,
std::char_traits<char> >::ostream_type {aka std::basic_ostream<char>}' and
'const std::pair<std::__cxx11::basic_string<char>,
std::__cxx11::basic_string<char> >')
*_M_out << __value;
有人知道为什么吗?
巴里(Barry)对这个问题给出了正确答案.但是,它不能解决问题,并且运行手动循环并不意味着重用现有的stl代码,因此问题被扩展到:
Barry has given the right answer to the question. It however does not solve the problem, and running a manual loop is not in the sense of reusing existing stl code, so the question gets extended to:
是否可以使流运算符在不污染std名称空间的情况下工作?
Is it possible to make the stream operator work without polluting the std namespace?
推荐答案
在ostream_joiner
实现的某个地方,将尝试进行以下操作:
Somewhere inside of the implementation of ostream_joiner
, there will be an attempt to something like:
os << value;
其中os
是std::basic_ostream
,值是您的pair
类型.为了确定该operator<<
调用的操作,我们查找了在此模板定义时可见的所有重载operator<<()
以及该
where os
is a std::basic_ostream
and value is your pair
type. In order to determine what to do for that operator<<
call, we lookup all the overloads operator<<()
visible at the point of definition of this template as well as as the overloads in the associated namespaces of the arguments (this is known as argument-dependent lookup).
当您使用您的 struct pair
时,pair
的关联命名空间为::
,因此ADL会找到您的::operator<<(std::ostream&, pair const&)
.这种超载有效,可以选择,一切都会高兴.
When you use your struct pair
, the associated namespace of pair
is ::
, so ADL will find your ::operator<<(std::ostream&, pair const&)
. This overload works, is chosen, everything is happy.
当您使用std::pair
时,pair
的关联命名空间是std
,并且找不到使用std::pair
的operator<<()
.因此,错误.
When you use std::pair
, the associated namespace of pair
is std
and there is no operator<<()
that can be found that takes a std::pair
. Hence the error.
您可以改为在自己的命名空间中创建自己的类型,可以为其添加重载的operator<<
,这可以完全是您自己的类型(问题中的方式),也可以继承
You could instead create your own type in your own namespace for which you can add an overloaded operator<<
, this can be fully your own type (the way it is in the question) or you could inherit the one in std
:
struct pair : std::pair<string,string> {
using std::pair<string,string>::pair;
};
std::ostream& operator<<(std::ostream&, my_pair const& ) {...}
或者,您不能使用make_ostream_joiner
.可以替换为:
Alternatively, you could just not use make_ostream_joiner
. Could replace this:
std::copy(std::begin(pairs),
std::end(pairs),
std::experimental::make_ostream_joiner(std::cout, ", "));
与此:
const char* delim = "";
for (auto const& pair : pairs) {
std::cout << delim << pair; // now, our point of definition does include
// our operator<<() declaration, we don't need ADL
delim = ", ";
}
这篇关于std :: experimental :: ostream_joiner和std :: pair的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!