我从 git 获得了 MSVC 的 range-v3。并通过MSVC C++14编译器编译。
考虑代码:
auto getter2 = [](const std::string&r) { return r+r; };
std::vector<std::string> vv = { "11","22", "33" };
std::cout << (vv | view::transform(getter2) | action::join) << std::endl;
它按预期工作。 但是我想添加分隔符加入,如果我写
std::cout << (vv | view::transform(getter2) | action::join(",")) << std::endl;
代码未编译。怎么了? 大量编译器错误如下:
1>d:\sources\ranges_test\ranges_test.cpp(28): error C2672: 'operator __surrogate_func': no matching overloaded function found
1>d:\sources\ranges_test\ranges_test.cpp(28): error C2783: 'unknown-type ranges::v3::action::action<ranges::v3::action::join_fn>::operator ()(Rng &,Rest &&...) const': could not deduce template argument for '__formal'
1>d:\sources\fingrad\dev.fingrad\src\vc\lib\range\v3\action\action.hpp(120): note: see declaration of 'ranges::v3::action::action<ranges::v3::action::join_fn>::operator ()'
1>d:\sources\ranges_test\ranges_test.cpp(28): error C2893: Failed to specialize function template 'unknown-type ranges::v3::action::action<ranges::v3::action::join_fn>::operator ()(T &&,Rest &&...) const'
1>d:\sources\ranges_test\ranges_test.cpp(28): note: With the following template arguments:
1>d:\sources\ranges_test\ranges_test.cpp(28): note: 'T=const char (&)[2]'
1>d:\sources\ranges_test\ranges_test.cpp(28): note: 'Rest={}'
1>d:\sources\ranges_test\ranges_test.cpp(28): note: 'A=ranges::v3::action::join_fn'
更新
不仅 MSVC 端口行为如此,here也是
最佳答案
猜测:字符串文字不是范围。尝试将 std::string(",")
或 view::c_str(",")
传递给 view::join
。
更新:啊。试试这个(用 clang trunk 测试):
#include <range/v3/all.hpp>
#include <vector>
#include <iostream>
int main()
{
using namespace ranges;
std::vector<std::string> vv = { "11","22", "33" };
auto getter2 = [](auto&r) { return view::concat(r, r); };
std::cout << (vv | view::transform(getter2)
| view::join(',')
| to_<std::string>()) << std::endl;
return 0;
}
关于c++ - range-v3 如何操作::使用定界符加入,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48383671/