Class/struct invariants

bearophile bearophileHUGS at
Wed Jun 16 05:08:53 PDT 2010

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.


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