Re: [boost] [convert] no-throw and fallback feature

Subject: Re: [boost] [convert] no-throw and fallback feature
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2011-05-10 14:37:47

Vicente BOTET wrote:
>
> > For general use with optional, I think the opt_tie() overload
> > might be useful.
> >
> Sorry Robert. You lost me.

I'm sorry.  I'll try again.

> Are you using the word overload instead of conversion?

No.

Compare the following options:

// ------------------------
// 1
int i(fallback);
optional<int> const o(convert_to<optional<int>>(s));
if (o)
{
   i = *o;
}
else
{
   std::cerr << "using fallback\n";
}

// ------------------------
// 2
int i(fallback);
bool b;
opt_tie(i, b) = convert_to<optional<int>>(s);
if (!b)
{
   std::cerr << "using fallback\n";
}

// ------------------------
// 3 (Vladimir dislikes this one)
int i(fallback);
if (!try_convert_to(s, i))
{
   std::cerr << "using fallback\n";
}

Notice that, for those who dislike try_convert_to(), #2 is nicer than #1.

Of course, if reporting the use of the fallback isn't needed, and the fallback isn't expensive to compute, there are these options:

// ------------------------
// A
int i(fallback);
opt_tie(i) = convert_to<optional<int>>(s);

// ------------------------
// B
int const i(convert_to<optional<int>>(s).get_value_or(fallback));

// ------------------------
// C
int i(fallback);
try_convert_to(s, i);

For those who dislike try_convert_to(), A is a more convenient option than B.

Both overloads of opt_tie() are useful.  Both assign to their first argument, if the optional is set.  The one with the extra argument also sets the optional's state in that argument.  That's easy enough to explain and use.

_____
Rob Stewart                           robert.stewart_at_[hidden]
Software Engineer                     using std::disclaimer;
Dev Tools & Components
Susquehanna International Group, LLP  http://www.sig.com





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