• bearophile (34/34) Jun 15 2010 Are D invariants supposed to be so "relaxed"? They don't get called with...
• Steven Schveighoffer (14/47) Jun 16 2010 Default construction for structs is a weird animal in D. A struct can
• bearophile (14/21) Jun 16 2010 Thank you for your answers, I was trying to understand.
• bearophile (1/1) Jun 16 2010 http://d.puremagic.com/issues/show_bug.cgi?id=4331
• Simen kjaeraas (5/12) Jun 16 2010 Ah, but I do. If it is an error to create an uninitialized struct of
• Steven Schveighoffer (7/19) Jun 16 2010 Yes, but an invariant doesn't guarantee that, since it is non-existent i...
• bearophile (4/6) Jun 16 2010 It's a good idea. I just hope this disabling will have an explicit & rea...
• Stewart Gordon (4/5) Jun 17 2010
• bearophile (4/5) Jun 17 2010 Thank you, then my bug 4329 is a dupe, I'll mark it so :-)

bearophile <bearophileHUGS lycos.com> writes:

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

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

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

bearophile <bearophileHUGS lycos.com> writes:

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

bearophile <bearophileHUGS lycos.com> writes:

http://d.puremagic.com/issues/show_bug.cgi?id=4331

"Simen kjaeraas" <simen.kjaras gmail.com> writes:

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

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

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

bearophile <bearophileHUGS lycos.com> writes:

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

Stewart Gordon <smjg_1998 yahoo.com> writes:

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.

bearophile <bearophileHUGS lycos.com> writes:

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