This is a part of an autopointer template i'm using. I originally had a
Get() member that returned the pointer, but thought it would nice to be able
to use the -> operator.
The question is why does the following line work, and call
cAP.m_pcPointer->Testing() ?
cAP->Testing(); // ???
cAP-> returns a CTest*, so it looks to me like the line reads
cAP.m_pcPointer <whitespace> Testing(), but it actually does
cAP.m_pcPointer->Testing(). I thought maybe it would be called with a bad
this-pointer, but it's the this-pointer is correct inside Testing(). What am
I not seeing here ?
template <class T> class TAutoPointer
{
private:
T
m_pcPointer;
public:
TAutoPointer(T t) : m_pcPointer(t) {}
~TAutoPointer() {if (m_pcPointer) delete m_pcPointer;}
T& operator -> ()
{
return m_pcPointer;
}
};
class CTest
{
int
m_nTest;
public:
CTest() : m_nTest(123) {}
void Testing()
{
int
i = 0;
}
};
typedef TAutoPointer < CTest* > CTestAutoPtr;
int main(int argc, char* argv[])
{
CTestAutoPtr
cAP(new CTest);
cAP->Testing();
return 0;
} 4 1780
PKH wrote: This is a part of an autopointer template i'm using. I originally had a Get() member that returned the pointer, but thought it would nice to be able to use the -> operator. The question is why does the following line work, and call cAP.m_pcPointer->Testing() ?
cAP->Testing(); // ???
cAP-> returns a CTest*, so it looks to me like the line reads cAP.m_pcPointer <whitespace> Testing(), but it actually does cAP.m_pcPointer->Testing(). I thought maybe it would be called with a bad this-pointer, but it's the this-pointer is correct inside Testing(). What am I not seeing here ?
You're not making a difference between the operator-> and the x->y
statement. cAp->Testing() will first execute operator-> for cAp and then
call Testing on the returned pointer.
If you want to do that explicitly, you can write:
cAP.operator->()->Testing();
PKH wrote:
[snip] The question is why does the following line work, and call cAP.m_pcPointer->Testing() ?
cAP->Testing(); // ???
cAP-> returns a CTest*, so it looks to me like the line reads cAP.m_pcPointer <whitespace> Testing(), but it actually does cAP.m_pcPointer->Testing(). I thought maybe it would be called with a bad this-pointer, but it's the this-pointer is correct inside Testing(). What am I not seeing here ?
The returned pointer from operator-> is automatically dereferenced
for you in any case. So operator-> is a little bit special, as
it doesn't return the final 'result' of the operation, but it's
result is used for an additional operation. Same with operator*()
Without that behaviour it would not be possible to use those
operators without major syntax changes.
--
Karl Heinz Buchegger kb******@gascad .at
"PKH" <no************ @online.no> schrieb im Newsbeitrag
news:19******** ************@ne ws4.e.nsc.no... This is a part of an autopointer template i'm using. I originally had a Get() member that returned the pointer, but thought it would nice to be
able to use the -> operator. The question is why does the following line work, and call cAP.m_pcPointer->Testing() ?
cAP->Testing(); // ???
cAP-> returns a CTest*, so it looks to me like the line reads cAP.m_pcPointer <whitespace> Testing(), but it actually does cAP.m_pcPointer->Testing(). I thought maybe it would be called with a bad this-pointer, but it's the this-pointer is correct inside Testing(). What
am I not seeing here ?
template <class T> class TAutoPointer { private: T m_pcPointer;
public: TAutoPointer(T t) : m_pcPointer(t) {} ~TAutoPointer() {if (m_pcPointer) delete m_pcPointer;}
T& operator -> () { return m_pcPointer; } };
class CTest { int m_nTest;
public: CTest() : m_nTest(123) {}
void Testing() { int i = 0; } };
typedef TAutoPointer < CTest* > CTestAutoPtr; int main(int argc, char* argv[]) { CTestAutoPtr cAP(new CTest);
cAP->Testing();
return 0; }
The magic of operator -> !
operator -> is an unary (!) operator, and it's overloading has a very
special meaning:
an overloaded operator -> must return something to that another operator ->
can be applied.
An example for illustration:
struct S { int i; }
S s = { 4711 };
class X
{
public:
S* operator->() { return &s; }
};
class Y
{
public:
X operator->() { return X(); }
};
class Z
{
public:
Y operator->() { return Y(); }
};
Z z;
int v = z->i;
Here, v will be initialized with 4711!
Why? "z->" results to an object of type Y, which has an operator->. This
operator results to an X with an operator-> resulting an pointer S*. To this
pointer the simple C++ operator-> for pointers can be applied and gives the
value of s.i .
So
int v = z->i;
will be compiled as
int v = z.Z::operator->().Y::operat or->().X::operat or->()->i;
In your code cAP-> results to an CTest*& to which the operator -> for
pointers will be applied and call CTest::Testing( ).
Greeting from Cologne, Germany
Georg
Ok, thanks for clearing that up. I didn't know the operator worked like
that.
PKH
"Georg Krichel" <so***@no.spa m> wrote in message
news:bP******** ******@s080a106 0.group.rwe.com ... "PKH" <no************ @online.no> schrieb im Newsbeitrag news:19******** ************@ne ws4.e.nsc.no... This is a part of an autopointer template i'm using. I originally had a Get() member that returned the pointer, but thought it would nice to be able to use the -> operator. The question is why does the following line work, and call cAP.m_pcPointer->Testing() ?
cAP->Testing(); // ???
cAP-> returns a CTest*, so it looks to me like the line reads cAP.m_pcPointer <whitespace> Testing(), but it actually does cAP.m_pcPointer->Testing(). I thought maybe it would be called with a bad this-pointer, but it's the this-pointer is correct inside Testing(). What am I not seeing here ?
template <class T> class TAutoPointer { private: T m_pcPointer;
public: TAutoPointer(T t) : m_pcPointer(t) {} ~TAutoPointer() {if (m_pcPointer) delete m_pcPointer;}
T& operator -> () { return m_pcPointer; } };
class CTest { int m_nTest;
public: CTest() : m_nTest(123) {}
void Testing() { int i = 0; } };
typedef TAutoPointer < CTest* > CTestAutoPtr; int main(int argc, char* argv[]) { CTestAutoPtr cAP(new CTest);
cAP->Testing();
return 0; }
The magic of operator -> !
operator -> is an unary (!) operator, and it's overloading has a very special meaning: an overloaded operator -> must return something to that another operator -> can be applied.
An example for illustration:
struct S { int i; } S s = { 4711 };
class X { public: S* operator->() { return &s; } };
class Y { public: X operator->() { return X(); } };
class Z { public: Y operator->() { return Y(); } };
Z z; int v = z->i;
Here, v will be initialized with 4711! Why? "z->" results to an object of type Y, which has an operator->. This operator results to an X with an operator-> resulting an pointer S*. To this pointer the simple C++ operator-> for pointers can be applied and gives the value of s.i . So int v = z->i; will be compiled as int v = z.Z::operator->().Y::operat or->().X::operat or->()->i;
In your code cAP-> results to an CTest*& to which the operator -> for pointers will be applied and call CTest::Testing( ).
Greeting from Cologne, Germany Georg
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