Can someone clarify the rules governing conversion functions?  I have
written a String class (who hasn't?) which has as the following constructor:

    String :: String (char s*);

and a member function strstr() with the following declaration:

    size_t String :: strstr (String& pattern);

Everything works fine when I call strstr() with a String argument, but
when I call it with a char * argument, such as:

    String s = "Test String";
    size_t pos = s.strstr ("Str");

I get a nasty message from the compiler (Borland C++ 2.0) that reads:

'Type mismatch in parameter 'pattern' in call to 'String::strstr(String near&)'

I have examined the sections in the ARM concerning conversion functions
(12.3) and function argument matching (13.2) and I don't understand why
the compiler doesn't convert the char * argument to a temporary String
object, since a constructor is declared with exactly one argument of
type char *.  Can someone clarify what is happening here?  The only way
around this problem that I have come up with is to redefine the argument
to strstr() as char * and then define an explicit "operator char* ()" con-
version function to convert String parameters to char *.  Although this
works, it defeats the purpose (and some of the functionality) of the class.

Thanks,

Dave.

In article <1644@caslon.cs.arizona.edu>, dwebster@cs.arizona.edu (Dave E. Webster, Jr.) writes:
>
>     ...  I have ... a String class ... which has ... :

>     String :: String (char s*);
  ...
>     size_t String :: strstr (String& pattern);


> when I call [strstr()] with a char * argument, such as:

>     String s = "Test String";
>     size_t pos = s.strstr ("Str");

> I get a nasty message from the compiler (Borland C++ 2.0) that reads:

> 'Type mismatch in parameter 'pattern' in call to 'String::strstr(String near&)'

> ... I don't understand why the compiler doesn't convert the char * argument
> to a temporary String object, since a constructor is declared with exactly
> one argument of type char *.  Can someone clarify what is happening here?

You have two problems of the same sort.  The double-quoted string
literal is treated as  const char* , not  char* .  And your  strstr()
needs a reference to  String , so C++ assumes that it may modify the
argument and refuses (or will refuse, in ANSI C++) to insert a conversion
that requires a temporary.  Your  strstr()  should accept an argument of
type  const String& .

C++ requires attention to  const .  It's good practice anyway, but C++
gives it a payoff.
--

 (This man's opinions are his own.)
 From mole-end    Mark Terribile

In article <1644@caslon.cs.arizona.edu> dwebster@cs.arizona.edu (Dave E. Webster, Jr.) writes:
>Everything works fine when I call strstr() with a String argument, but
>when I call it with a char * argument, such as:
>
>    String s = "Test String";
>    size_t pos = s.strstr ("Str");
>
>I get a nasty message from the compiler (Borland C++ 2.0) that reads:
>
>'Type mismatch in parameter 'pattern' in call to 'String::strstr(String near&)'
>
 It's a bug.  The workaround is to explicitly cast the quoted string
to a String:

 size_t pos = s.strstr( String("Str) );

 -- Pete