How do i use this unique operator overloaded function

Question

I came across this operator overload during my use of the sqlapi which does exactly what i need, but i'm not sure how it works.

sqlapi.h

class SQLAPI_API SAString
{
public:
//stuff

//! Return pointer to const string
operator const SAChar *() const;

//stuff
}

The only way that i know how to use it is because of the intellisense which doesn't look pretty:

char* chPointer = SAStringObj.operator const SAChar*();

Questions:

Is there a better looking way to call this overloaded operator?

Can someone dissect what this function header tells us about the function?

"*operator const SAChar () const;"

`SAString myString; const SAChar* chPointer = myString;` – Igor Tandetnik May 10 '15 at 00:48 — Igor Tandetnik, May 10 '15 at 00:48

score 3 · Accepted Answer · edited May 23 '17 at 12:30

This is an implicit cast operator for converting an SAString to const SAChar *. It is implicitly invoked whenever you use an SAString in a context in which a const SAChar * is expected (unless it would be ambiguous).

void foo(const SAChar *);

SAString myString("bar");

foo(myString);
const SAChar *myCharPtr = myString;

You can use a static_cast to force it in other contexts:

static_cast<const SAChar *>(myString)

You can use boost::implicit_cast which is safer:

boost::implicit_cast<const SAChar *>(myString)

In C++11 you can have an explicit cast operator which is only valid in explicit cast contexts:

explicit operator const SAChar *() const;

The const at the end has the usual meaning, just as for ordinary member functions.

Wow, that answer was perfect and answered all my questions. Also, is there any difference between a standard "(const SAChar*)myString" cast and a static_cast? — shawn a, May 10 '15 at 01:25

How do i use this unique operator overloaded function

1 Answers1