如何在元组初始化向量中删除样板?

我想初始化一个 5 元组字符串的列表，如下所示：

std::vector<std::tuple<std::string,std::string,std::string,std::string,std::string> >
examples = 
{
{"/foo"                 ,"/"           ,"foo"        ,""   ,"foo"},
{"/foo/"                ,"/"           ,"foo"        ,""   ,"foo"},
{"/foo//"               ,"/"           ,"foo"        ,""   ,"foo"},
{"/foo/./"              ,"/foo"        ,"."          ,""   ,""},
{"/foo/bar"             ,"/foo"        ,"bar"        ,""   ,"bar"},
{"/foo/bar."            ,"/foo"        ,"bar."       ,""   ,"bar"},
{"/foo/bar.txt"         ,"/foo"        ,"bar.txt"    ,"txt","bar"},
{"/foo/bar.txt.zip"     ,"/foo"        ,"bar.txt.zip","zip","bar.txt"},
{"/foo/bar.dir/"        ,"/foo"        ,"bar.dir"    ,"dir","bar"},
{"/foo/bar.dir/file"    ,"/foo/bar.dir","file"       ,""   ,"file"},
{"/foo/bar.dir/file.txt","/foo/bar.dir","file.txt"   ,"txt","file"}
};

这个问题问为什么嵌套初始值设定项列表不能用于元组向量：答案说使用std::make_tuple。但这让我的代码看起来很荒谬：

std::vector<std::tuple<std::string,std::string,std::string,std::string,std::string> >
examples = 
{
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo"                 ,"/"           ,"foo"        ,""   ,"foo"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/"                ,"/"           ,"foo"        ,""   ,"foo"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo//"               ,"/"           ,"foo"        ,""   ,"foo"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/./"              ,"/foo"        ,"."          ,""   ,""),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar"             ,"/foo"        ,"bar"        ,""   ,"bar"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar."            ,"/foo"        ,"bar."       ,""   ,"bar"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar.txt"         ,"/foo"        ,"bar.txt"    ,"txt","bar"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar.txt.zip"     ,"/foo"        ,"bar.txt.zip","zip","bar.txt"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar.dir/"        ,"/foo"        ,"bar.dir"    ,"dir","bar"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar.dir/file"    ,"/foo/bar.dir","file"       ,""   ,"file"),
std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo/bar.dir/file.txt","/foo/bar.dir","file.txt"   ,"txt","file")
};

如果我无法摆脱std::make_tuple<...>，我至少可以使用typedef或using来消除代码中的混乱吗？

using StringQuintet = std::tuple<std::string,std::string,std::string,std::string,std::string>;没有帮助，因为std::make_tuple<...>只需要元组模板参数而不是元组类型。

有没有一个好方法可以清理这个样板看起来的烂摊子？