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June 16, 2010 Class/struct invariants | ||||
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Are D invariants supposed to be so "relaxed"? They don't get called with default constructors: struct Foo { int x = 0; this(int xx) { this.x = xx; } invariant() { assert(x == 1); } } struct Bar { int x = 0; invariant() { assert(x == 1); } } class CFoo { int x = 0; this(int xx) { this.x = xx; } invariant() { assert(x == 1); } } class CBar { int x = 0; invariant() { assert(x == 1); } } void main() { Foo f1; // no asserts Foo f2 = Foo(); // no asserts // Foo f3 = Foo(0); // asserts, good Bar b1; // no asserts Bar b2 = Bar(); // no asserts Bar b3 = Bar(0); // no asserts //assert(b3); // can't be used // b3.__invariant(); // asserts //CFoo f3 = new CFoo(0); // asserts, good CBar cb2 = new CBar(); // no asserts //assert(cb2); // asserts, good } Bye, bearophile |
June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to bearophile | On Tue, 15 Jun 2010 21:29:27 -0400, bearophile <bearophileHUGS@lycos.com> wrote:
> Are D invariants supposed to be so "relaxed"? They don't get called with default constructors:
>
>
> struct Foo {
> int x = 0;
> this(int xx) { this.x = xx; }
> invariant() { assert(x == 1); }
> }
> struct Bar {
> int x = 0;
> invariant() { assert(x == 1); }
> }
> class CFoo {
> int x = 0;
> this(int xx) { this.x = xx; }
> invariant() { assert(x == 1); }
> }
> class CBar {
> int x = 0;
> invariant() { assert(x == 1); }
> }
> void main() {
> Foo f1; // no asserts
> Foo f2 = Foo(); // no asserts
> // Foo f3 = Foo(0); // asserts, good
>
> Bar b1; // no asserts
> Bar b2 = Bar(); // no asserts
> Bar b3 = Bar(0); // no asserts
>
> //assert(b3); // can't be used
> // b3.__invariant(); // asserts
>
> //CFoo f3 = new CFoo(0); // asserts, good
>
> CBar cb2 = new CBar(); // no asserts
> //assert(cb2); // asserts, good
> }
Default construction for structs is a weird animal in D. A struct can always be default constructed and is always initialized to s.init. This allows you to construct for instance an array of structs with simple memory copying.
During default struct construction, no constructors are run (they aren't allowed anyways) and no invariants are run. What would be the point of running an invariant during default construction? The only think it could possibly do is make code like this:
S s;
Fail without -release, and pass with -release. I don't see the value in that.
-Steve
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June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to Steven Schveighoffer | Steven Schveighoffer:
> During default struct construction, no constructors are run (they aren't
> allowed anyways) and no invariants are run. What would be the point of
> running an invariant during default construction? The only think it could
> possibly do is make code like this:
> S s;
> Fail without -release, and pass with -release. I don't see the value in
> that.
Thank you for your answers, I was trying to understand.
Of all the examples I have shown this can be the worst:
struct Foo {
int x;
invariant() { assert(x > 0); }
}
void main() {
Foo f = Foo(-10);
}
Here I'd like the compiler to assert (at compile time or at runtime), or to refuse an invariant in structs like that, where I think D has no way to enforce it (unless you call __invariant(), but this is silly).
Later I can write a "bug" report about this.
Bye,
bearophile
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June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to Steven Schveighoffer | Steven Schveighoffer <schveiguy@yahoo.com> wrote: > During default struct construction, no constructors are run (they aren't allowed anyways) and no invariants are run. What would be the point of running an invariant during default construction? The only think it could possibly do is make code like this: > > S s; > > Fail without -release, and pass with -release. I don't see the value in that. Ah, but I do. If it is an error to create an uninitialized struct of type S, then the above code is a bug, is it not? -- Simen |
June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to Simen kjaeraas | On Wed, 16 Jun 2010 10:58:28 -0400, Simen kjaeraas <simen.kjaras@gmail.com> wrote:
> Steven Schveighoffer <schveiguy@yahoo.com> wrote:
>> During default struct construction, no constructors are run (they aren't allowed anyways) and no invariants are run. What would be the point of running an invariant during default construction? The only think it could possibly do is make code like this:
>>
>> S s;
>>
>> Fail without -release, and pass with -release. I don't see the value in that.
>
> Ah, but I do. If it is an error to create an uninitialized struct of
> type S, then the above code is a bug, is it not?
Yes, but an invariant doesn't guarantee that, since it is non-existent in release mode, and a compile time error is better.
What you want is to be able to disable the default constructor. Andrei has hinted it might be a future improvement on other threads.
-Steve
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June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to Steven Schveighoffer | Steven Schveighoffer:
> What you want is to be able to disable the default constructor. Andrei has hinted it might be a future improvement on other threads.
It's a good idea. I just hope this disabling will have an explicit & readable syntax (maybe @disable can be used here).
Bye,
bearophile
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June 16, 2010 Re: Class/struct invariants | ||||
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Posted in reply to bearophile | http://d.puremagic.com/issues/show_bug.cgi?id=4331 |
June 17, 2010 Re: Class/struct invariants | ||||
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Posted in reply to bearophile | bearophile wrote: > Are D invariants supposed to be so "relaxed"? They don't get called with default constructors: <snip> http://d.puremagic.com/issues/show_bug.cgi?id=519 Stewart. |
June 17, 2010 Re: Class/struct invariants | ||||
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Posted in reply to Stewart Gordon | Stewart Gordon:
> http://d.puremagic.com/issues/show_bug.cgi?id=519
Thank you, then my bug 4329 is a dupe, I'll mark it so :-)
Bye,
bearophile
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