Re: [boost] Variadic append for std::string

23 Jan 2017

      Hi,

Am 23.01.2017 16:32, schrieb Hans Dembinski:
...
...
On 23 Jan 2017, at 11:23, Christof Donat  wrote:
auto my_new_str =
   concat("Hello ",
          join(", ",std::begin(my_nums), std::end(my_nums)),
          format(" the file %1% contains %2% bytes", filename, 
filesize)).str();
It makes sense to me for "join" to return a string factory, because it
is likely to be nested in "concat". But I don't see the practical case
of nested "concat" calls, at least it is not going to be a common
pattern in the need of optimising.
There is several usecases:

1. scope for formatting tags:

concat(format::hex<int>, 42, " is hex for ", concat(42)).str();

Here the inner concat will convert the 42 to its decimal representation, 
while the outer one converts the first 42 to its hex representation.

2. concat() in calls to format():

format("%|1$40t|%2%", concat(first_name, " ", last_name), 
phone_number).str();

format().str() will allocate the buffer and ask the concat string 
factory to write into it.

3. results from concat() in a boost::range that is passed to join():

join(separator("\n"),
      my_files | transformed([](const std::filesystem::path& f) -> auto {
                                 return concat(f.filename, ": ", 
std::filesystem::file_size(f));
                             })).str();

join().str() will ask every concat string factory to render directly 
into the common buffer.
...
If "concat" is the outer layer anyway, I would return a std::string
directly for convenience. It is easy to forget the trailing .str() and
it does not look elegant.
Of course better proposals are welcome :-) Would you prefer the implicit 
conversion? If so, why?

Christof