How to remove boilerplate in vector of tuple initialization?

Question

I'd like to initialize a list of 5-tuples of strings like this:

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"}
  };

This question asks why nested initializer lists can't be used for vector of tuples: the answers say to use std::make_tuple. But this makes my code look ridiculous:

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")
  };

If I can't get rid of the std::make_tuple<...>, can I at least use typedef or using to remove the clutter in the code?

using StringQuintet = std::tuple<std::string,std::string,std::string,std::string,std::string>; doesn't help because std::make_tuple<...> needs just the tuples template arguments not the tuple type.

Is there a good way to clean up this boilerplate looking mess?

Using explicit template arguments in `std::make_tuple` defeats its purpose. Just use `std::tuple` directly. — tkausl, Sep 26 '18 at 02:57
the point of `make_tuple` is to deduce types from the initializers, so you don't have to specify the type list — M.M, Sep 26 '18 at 02:58
std::make_tuple() is a function it will deduce the types so you don't need to explicitly mention them. Though you will need to use the `std::string` literal rather than the `C-String` literal. — Martin York, Sep 26 '18 at 03:03
Everyone is mentioning that make_tuple doesn't need the template arguments. This is great! (missed that) I will happily accept or just close... — Alec Jacobson, Sep 26 '18 at 03:04
Your original code should work with a C++17 compiler because `std::string` can be constructed implicitly from a `char const *`, see https://stackoverflow.com/a/32084829/241631 — Praetorian, Sep 26 '18 at 03:21

score 3 · Accepted Answer · answered Sep 26 '18 at 03:06

 std::make_tuple<std::string,std::string,std::string,std::string,std::string>("/foo"                 ,"/"           ,"foo"        ,""   ,"foo"),

you don't pass types to make tuple.

std::make_tuple("/foo"                 ,"/"           ,"foo"        ,""   ,"foo"),

you let the compiler deduce them. Alternatively:

std::make_tuple("/foo"s                 ,"/"s           ,"foo"s        ,""s   ,"foo"s),

to have std string literals (using namespace std::literals; to permit this).

Or:

using StringQuintet = std::tuple<std::string,std::string,std::string,std::string,std::string>;
// ...
StringQuintet("/foo"                 ,"/"           ,"foo"        ,""   ,"foo"),

score 0 · Answer 2 · answered Sep 26 '18 at 03:05

0

Use std::make_tuple:

using namespace std::string_literals;
....

std::make_tuple("/foo"s                 ,"/"s           ,"foo"s        ,""s   ,"foo"s),

....

answered Sep 26 '18 at 03:05

Martin York

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How to remove boilerplate in vector of tuple initialization?

2 Answers2