• Jay (92/92) Nov 13 2004 Hi,
• Sean Kelly (5/12) Nov 13 2004 This is correct behavior. It has to do with the symbol lookup rules. T...

Jay <Jay_member pathlink.com> writes:

Hi, 
 
In the following program, there is a base class with an accessor and a
modifier overloading the same name, to function like a class `property'. There
are two classes derived from the base, one is trivial, and the other overrides
the base modifier. The former compiles fine, whereas the latter causes the
compiler to think that there is no base accessor: see embedded comment below.
Use -version=BROKEN or -fversion=BROKEN on the command line to observe the
compiler error. 
 
The language documentation says: 
 
`In D, function
overloading is simple. It matches exactly, it matches with 
implicit
conversions, or it does not match. If there is more than one match, 
it is an
error.' 
 
I don't think the above is the cause of the problem, but I'm only two
D-weeks 
old and I may be wrong. Am I? 
 
Thanks, 
Jay
-------------------------------------- 
 
class Base 
{ 
  public: 
    char[]
prop() { return _prop; } 
    void prop(char[] v) { _prop = v; } 
  private:
char[] _prop; 
} 
 
class Derived1 : public Base 
{ 
  public: 
    void print()
{ 
        printf("Derived1.prop = %*s\n", prop);
printf("Derived1.prop() = %*s\n", prop()); 
    } 
} 
 
class Derived2 : public
Base 
{ 
  public: 
    override void prop(char[] v) 
    { 
        super.prop
= v; 
        if (_prop == "No17") { 
            printf("Special case\n");
} 
    } 
    void print() 
    { 
        version (BROKEN) { 
            //
The following two uses of `prop' should refer to 
            // Base.prop(),
shouldn't they?  The compiler wants 
            // them to be
Derived2.prop(char[]). 
            printf("Derived2.prop = %*s\n", prop);
printf("Derived2.prop() = %*s\n", prop()); 
        } else {
printf("Derived2.super.prop = %*s\n", super.prop); 
        } 
    } 
} 
 
int
main() 
{ 
    Derived1 d1 = new Derived1; 
    Derived2 d2 = new Derived2;
d1.prop = "No17"; 
    d1.prop = "No16"; 
    d1.print(); 
 
    d2.prop =
"No17"; 
    d2.prop = "No16"; 
    d2.print(); 
 
    return 0; 
} 

Sean Kelly <sean f4.ca> writes:

In article <cn4m29$dna$1 digitaldaemon.com>, Jay says...
Hi, 
 
In the following program, there is a base class with an accessor and a
modifier overloading the same name, to function like a class `property'. There
are two classes derived from the base, one is trivial, and the other overrides
the base modifier. The former compiles fine, whereas the latter causes the
compiler to think that there is no base accessor: see embedded comment below.

This is correct behavior.  It has to do with the symbol lookup rules.  To get
everything working as you want, add an alias for the accessor in the derived
class.


Sean