are scope guards (scope(exit, success, failure)) zero-cost abstractions?

Daniel Kozak kozzi11 at gmail.com
Thu Feb 8 11:23:43 UTC 2018


I mean scope(success), for scope(exit) there is no speed penalty

On Thu, Feb 8, 2018 at 12:03 PM, Daniel Kozak <kozzi11 at gmail.com> wrote:

> Yes, it add, but is almost zero
>
> On Thu, Feb 8, 2018 at 12:00 PM, Timothee Cour via Digitalmars-d-learn <
> digitalmars-d-learn at puremagic.com> wrote:
>
>> I know that, my question is whether it adds any runtime overhead over
>> naive way (which is to call the "bar" finalizer before each return
>> statement)  in the case where no exception is thrown
>>
>>
>> On Thu, Feb 8, 2018 at 2:44 AM, Mike Parker via Digitalmars-d-learn
>> <digitalmars-d-learn at puremagic.com> wrote:
>> > On Thursday, 8 February 2018 at 10:09:12 UTC, Timothee Cour wrote:
>> >>
>> >> I'm curious whether scope guards add any cost over the naive way, eg:
>> >>
>> >> ```
>> >> void fun(){
>> >>   ...
>> >>   scope(success) {bar;}
>> >>   ...
>> >> }
>> >> ```
>> >>
>> >> vs:
>> >>
>> >> ```
>> >> void fun(){
>> >>   ...
>> >>   if(foo1){
>> >>     bar;  // add this before each return
>> >>     return;
>> >>   }
>> >>   ...
>> >>   bar;
>> >>   return;
>> >> }
>> >> ```
>> >>
>> >> For scope(success) and scope(failure), the naive way would anyway
>> >> involve try/catch statements but what about scope(exit)? Does the
>> >> zero-cost exception model (zero cost being for non-thrown exceptions)
>> >> guarantee us that scope(success) has 0 overhead over naive way?
>> >
>> >
>> > Scope guards are lowered to the equivalent try/catch/finally blocks
>> anyway.
>>
>
>
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