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Subject: Re: [boost] Variadic append for std::string
From: Christof Donat (cd_at_[hidden])
Date: 2017-01-24 04:06:43
Hi,
Am 23.01.2017 19:26, schrieb Olaf van der Spek:
> On Mon, Jan 23, 2017 at 5:32 PM, Christof Donat <cd_at_[hidden]> wrote:
>> 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.
>
> Wouldn't concat(hex(42), " is hex for", 42) make more sense?
That is a valid approach for concat() and format(), but suboptimal for
join(). Think of this example:
join(hex<int>, separator(" "), my_nums);
Here all the numbers are converted to their hex representation. With
your approach this would look like:
join(separator(" "),
my_nums | transform([](int i) -> int {
return hex(i);
}));
That is much more difficult to understand.
Since for join() tag parameters to define the conversion is, to me, the
superior choice, I think, we should use it for concat() and format() as
well, for consistency.
>> 2. concat() in calls to format():
>>
>> format("%|1$40t|%2%", concat(first_name, " ", last_name),
>> phone_number).str();
>
> Why not fold the name concat into the format string?
In this example I want the full name to take up 40 characters, no matter
if the first name is long or short. With format strings as used by
boost::format I don't know how that could be achieved, and extending the
format language, of course, makes the interpreter slower and more
complex.
>>> 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?
>
> Implicit is problematic with auto..
That is one of the reasons, I prefer the explicit str() function. It
also fits well to other factory functions, we might prefer to have as
well, like e.g. concat(...).append_to(my_string), or
concat(...).overwrite(my_string).
Christof
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