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From: Robert Ramey (ramey_at_[hidden])
Date: 2019-06-27 17:12:12

On 6/27/19 9:47 AM, Andrey Semashev via Boost wrote:
> On 6/27/19 7:32 PM, Robert Ramey via Boost wrote:
>> On 6/27/19 8:43 AM, Andrey Semashev via Boost wrote:
>> - From the caller's perspective, a noexcept function will never throw.
>>>> In anycase I should have phrased the above as:
>>>> Is swap guaranteed to succeed?
>>> In general, no. It is only guaranteed to succeed if it is noexcept.
>>> See above when it is noexcept.
>> A... - I'm still confused.  How can one say that swap is guaranteed to
>> succeed when it can result in calling terminate()?
> First, people normally don't write noexcept functions knowing they might
> throw.

Right - but swap is a templated function. The obvious way to implement
swap is via moves on type T. So how could swap be guaranteed to succeed
for any T?

Meaning that the call to terminate() is equivalent to a bug.

LOL - right - I'm trying to figure out how to guarantee that my program
doesn't have a bug.

> Second, your code after a noexcept swap() returns will not be run if
> terminate() gets called.

> Meaning that the code is guaranteed to only
> operate when swap() successfully returned.
LOL - I get this. The question is: how can one know that swap will be
successfully returned? for a given type T.

In other words, how can it make sense that swap<T> be noexcept without
considering the specific type T?

Robert Ramey
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