Re: [boost] [review] string convert

Subject: Re: [boost] [review] string convert
From: Jeff Flinn (TriumphSprint2000_at_[hidden])
Date: 2011-05-05 13:20:21

Stewart, Robert wrote:
> Vicente BOTET wrote:
>> De : "Vladimir Batov"
>>> "Stewart, Robert"  wrote
>>> Maybe to avoid the temptation of pairing up we could simplify
>>> (at least from the user perspectives) down to:
>>>
>>> 1. T convert_cast(S)
>>> 2. T convert_cast(S, T, throw_t =nothrow)
>> I don't see why (2) the user could want to throw when is giving
>> a fallback. So the throw_t =nothrow is not needed.
> 
> That's the non-DefaultConstructible, no-sentinel use case:
> 
> struct example
> {
>    example(int);
> };
> 
> To get an exception on conversion failure, one writes:
> 
> example const e(convert_cast<example>("1")); // exception
> 
> However, without the default_value CP, that fails because example is non-DefaultConstructible.
> 
> One can avoid the exception by providing a fallback value:
> 
> example const e(convert_cast<example>("1", 3)); // no exception
> 
> (I'm assuming the fallback value can be of type U implicitly convertible to T.)
> 
> In that case, e == example(3).
> 
> But if one wants an exception to indicate conversion failure, an initial value is needed and the desire for an exception must be noted:
> 
> example const e(convert_cast<example>("1", 3, throw_)); // exception

Sorry if I'm just being dense, but if the above throws, the default 
value is never used, so why supply it?

> Obviously, with your default_value CP, the second argument isn't needed for the last use case:
> 
> template <>
> default_value<example>
> {
>    static example
>    apply()
>    {
>       return example(3);
>    }
> };
> 
> example const e(convert_cast<example>("1")); // exception

So it's really just the no-default function that needs to be told to 
throw or return an optional.

Jeff