| Topic: |
DEVELOP > c-Plus-Plus |
| User: |
"" |
| Date: |
28 Aug 2006 08:47:09 PM |
| Object: |
const does not apply to pointers? |
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
.
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| User: "Pete Becker" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 05:50:19 AM |
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wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
ch is const. The thing it points to is not. So you can change *ch, but
you can't change ch. Try it:
ch = 0; // illegal
.
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| User: "Pete Becker" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 05:53:56 AM |
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Pete Becker wrote:
grubertm@gmail.com wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
ch is const. The thing it points to is not. So you can change *ch, but
you can't change ch. Try it:
ch = 0; // illegal
Whoops, it's pChar that's const:
pChar = 0; // illegal
To prevent changing the char that it points to, make it a pointer to const:
const char *pChar;
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| User: "Pete Becker" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 06:24:19 AM |
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wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
Calling DoStuff doesn't change anything in your const object. pChar
points to something outside the object, and the code doesn't say that
the thing it points to shouldn't be modified. The type of
constClass.pChar is char *const, not const char *. To make it const
char*, define it that way:
const char *pChar;
.
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| User: "" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 01:13:48 PM |
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Thanks for all the feedback !
I was assuming that const'ing an object would basically prepend a
"const" to its members, as in:
char member1; becomes const char member1;
and
char* member2; becomes const char* member2;
Instead it can be visualized as appending the const:
char member1; becomes char const member1;
and
char* member2; becomes char* const member2;
The suggestion of adding a const in the class definition works for this
example but I can't use it in the real-life case where I ran into this
issue. Using accessors might be a viable solution though.
.
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| User: "Ron Natalie" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 01:21:33 PM |
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wrote:
Thanks for all the feedback !
I was assuming that const'ing an object would basically prepend a
"const" to its members, as in:
char member1; becomes const char member1;
and
char* member2; becomes const char* member2;
Instead it can be visualized as appending the const:
char member1; becomes char const member1;
and
char* member2; becomes char* const member2;
Actually it is NEITHER. It makes each member const.
You need to learn how declarations work or you're going
to be in for a world of grief.
What about
int (*member3)(char*, int);
???
The suggestion of adding a const in the class definition works for this
example but I can't use it in the real-life case where I ran into this
issue. Using accessors might be a viable solution though.
Huh? Do you know the difference between a pointer being const and
the object it points to being const?
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| User: "mlimber" |
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| Title: Re: const does not apply to pointers? |
30 Aug 2006 08:30:39 AM |
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wrote:
Thanks for all the feedback !
I was assuming that const'ing an object would basically prepend a
"const" to its members, as in:
char member1; becomes const char member1;
and
char* member2; becomes const char* member2;
Instead it can be visualized as appending the const:
char member1; becomes char const member1;
and
char* member2; becomes char* const member2;
The suggestion of adding a const in the class definition works for this
example but I can't use it in the real-life case where I ran into this
issue. Using accessors might be a viable solution though.
As others suggested, you can use other containers (e.g., std::vector)
that propogate constness to their containees. You can also implement a
pointer proxy class that propogates constness to its pointee. Something
like:
struct MyException {};
template <class T>
class DeepPtr
{
public:
DeepPtr( T* const ptr )
: m_ptr( ptr )
{
if( !ptr )
{
throw MyException(); // or whatever
}
}
// Standard access
T* operator->() { return m_ptr; }
T& operator*() { return *m_ptr; }
// Make pointee const if pointer is const
const T* operator->() const { return m_ptr; }
const T& operator*() const { return *m_ptr; }
private:
T * const m_ptr;
// Disabled copying
DeepPtr( const DeepPtr& );
DeepPtr& operator=( const DeepPtr& );
};
class A
{
DeepPtr<int> m_pi1;
int *const m_pi2
public:
A( int* const pi1, int* const pi2 )
: m_pi1( pi1 ), m_pi2( pi2 )
{
*m_pi1 = 0xEATBEEF;
*m_pi2 = 0xEATBEEF;
}
void Foo() const
{
*m_pi1 = 0xFEEDB0B; // Error!
*m_pi2 = 0xFEEDB0B; // Ok
}
void Foo()
{
*m_pi1 = 0xC0FFEE; // Ok
*m_pi2 = 0xC0FFEE; // Ok
}
};
The copy constructor and assignment operator are disable because they
would allow a backdoor to ignore constness (intentionally or not), and
as a result, this pointer proxy is not usable in standard containers
such as std::vector that require copy construction. It also means that
the example class A can't be copied either, but that may be acceptable
for your application.
Cheers! --M
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| User: "Earl Purple" |
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| Title: Re: const does not apply to pointers? |
31 Aug 2006 04:24:55 AM |
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mlimber wrote:
template <class T>
class DeepPtr
{
public:
DeepPtr( T* const ptr )
: m_ptr( ptr )
{
if( !ptr )
{
throw MyException(); // or whatever
}
I'd prefer to allow such pointers to be NULL. If not I'd rather use
asserts
but that's up to you.
}
// Standard access
T* operator->() { return m_ptr; }
T& operator*() { return *m_ptr; }
just a shame we can't overload operator .
Then we could make this a smart-reference (given that you don't want to
allow it to
be null or reassigned)
// Make pointee const if pointer is const
const T* operator->() const { return m_ptr; }
const T& operator*() const { return *m_ptr; }
I also might add methods to get the pointer other than operator-> When
you have a function that wants a (const) T* pointer you don't want to
call myPtr->operator->() to get it (or &*myPtr)
private:
T * const m_ptr;
// Disabled copying
DeepPtr( const DeepPtr& );
unnecessary to disable assignment
DeepPtr& operator=( const DeepPtr& );
};
class A
{
DeepPtr<int> m_pi1;
int *const m_pi2
public:
A( int* const pi1, int* const pi2 )
: m_pi1( pi1 ), m_pi2( pi2 )
{
*m_pi1 = 0xEATBEEF;
*m_pi2 = 0xEATBEEF;
// illegal: T is not a valid hex character,. You could assign to
0xEA7BEEF though.
}
The copy constructor and assignment operator are disable because they
would allow a backdoor to ignore constness (intentionally or not), and
as a result, this pointer proxy is not usable in standard containers
such as std::vector that require copy construction. It also means that
the example class A can't be copied either, but that may be acceptable
for your application.
The backdoor with copy-construction can be worked around by overloading
the copy-constructor to take a non-const reference (like auto_ptr
does). Then you can't duplicate a const "smart" pointer to remove its
constness.
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| User: "mlimber" |
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| Title: Re: const does not apply to pointers? |
31 Aug 2006 09:11:46 AM |
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Earl Purple wrote:
mlimber wrote:
template <class T>
class DeepPtr
{
public:
DeepPtr( T* const ptr )
: m_ptr( ptr )
{
if( !ptr )
{
throw MyException(); // or whatever
}
I'd prefer to allow such pointers to be NULL. If not I'd rather use
asserts
but that's up to you.
}
// Standard access
T* operator->() { return m_ptr; }
T& operator*() { return *m_ptr; }
just a shame we can't overload operator .
Then we could make this a smart-reference (given that you don't want to
allow it to
be null or reassigned)
// Make pointee const if pointer is const
const T* operator->() const { return m_ptr; }
const T& operator*() const { return *m_ptr; }
I also might add methods to get the pointer other than operator-> When
you have a function that wants a (const) T* pointer you don't want to
call myPtr->operator->() to get it (or &*myPtr)
Sure. As per Alexandrescu's suggestion in _MC++D_, I have it as a
friendly free-standing function in my real code since it's not an
"official" part of the interface.
private:
T * const m_ptr;
// Disabled copying
DeepPtr( const DeepPtr& );
unnecessary to disable assignment
It is necessary. In my production code, I have a copy constructor that
takes non-const, and this line exists just to make explicit that I did
not want to allow copying from const. Omitting this line without that
other copy constructor, however, can lead to the un/intentional
ignoring of const that I mentioned:
void Foo( const DeepPtr<int>& pConst )
{
DeepPtr<int> pNonConst( pConst );
*pConst = 42; // Error
*pNonConst = 42; // No error, but evil
}
DeepPtr& operator=( const DeepPtr& );
};
class A
{
DeepPtr<int> m_pi1;
int *const m_pi2
public:
A( int* const pi1, int* const pi2 )
: m_pi1( pi1 ), m_pi2( pi2 )
{
*m_pi1 = 0xEATBEEF;
*m_pi2 = 0xEATBEEF;
// illegal: T is not a valid hex character,. You could assign to
0xEA7BEEF though.
Typing too quickly again. How about 0xBADBEEF?
}
The copy constructor and assignment operator are disable because they
would allow a backdoor to ignore constness (intentionally or not), and
as a result, this pointer proxy is not usable in standard containers
such as std::vector that require copy construction. It also means that
the example class A can't be copied either, but that may be acceptable
for your application.
The backdoor with copy-construction can be worked around by overloading
the copy-constructor to take a non-const reference (like auto_ptr
does). Then you can't duplicate a const "smart" pointer to remove its
constness.
Precisely what I do in my real code. I was simplifying here so as not
to obscure the main issue.
Cheers! --M
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| User: "mlimber" |
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| Title: Re: const does not apply to pointers? |
31 Aug 2006 09:27:29 AM |
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mlimber wrote:
Earl Purple wrote:
The backdoor with copy-construction can be worked around by overloading
the copy-constructor to take a non-const reference (like auto_ptr
does). Then you can't duplicate a const "smart" pointer to remove its
constness.
Precisely what I do in my real code. I was simplifying here so as not
to obscure the main issue.
However, your point is well-taken, and I probably shouldn't have
bothered simplifying that part.
Cheers! --M
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| User: "Earl Purple" |
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| Title: Re: const does not apply to pointers? |
01 Sep 2006 10:09:25 AM |
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mlimber wrote:
Earl Purple wrote:
unnecessary to disable assignment
It is necessary.
Yes it is necessary that assignment cannot take place but not necessary
to explicitly disable it as the preference of a const member will
automatically do that for you.
.
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| User: "mlimber" |
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| Title: Re: const does not apply to pointers? |
01 Sep 2006 12:29:25 PM |
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Earl Purple wrote:
mlimber wrote:
Earl Purple wrote:
unnecessary to disable assignment
It is necessary.
Yes it is necessary that assignment cannot take place but not necessary
to explicitly disable it as the preference of a const member will
automatically do that for you.
Right you are. At least one of my compiler complains, however, if I
don't explicitly disable it.
Cheers! --M
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| User: "Noah Roberts" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 01:23:04 PM |
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wrote:
Thanks for all the feedback !
I was assuming that const'ing an object would basically prepend a
"const" to its members, as in:
char member1; becomes const char member1;
and
char* member2; becomes const char* member2;
Instead it can be visualized as appending the const:
char member1; becomes char const member1;
and
char* member2; becomes char* const member2;
The suggestion of adding a const in the class definition works for this
example but I can't use it in the real-life case where I ran into this
issue. Using accessors might be a viable solution though.
It is for reasons like this that I have stopped using prefix const
notation. Even though it speaks better, "a constant int pointer," it
creates inconsistancies such as this. So now I say, "an int that is
contstant pointer."
.
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| User: "Ron Natalie" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 01:27:31 PM |
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Noah Roberts wrote:
It is for reasons like this that I have stopped using prefix const
notation. Even though it speaks better, "a constant int pointer," it
creates inconsistancies such as this. So now I say, "an int that is
contstant pointer."
What? "an int that is a pointer?"
Ints aren't pointers.
How about a "pointer that is constant to int"
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| User: "Noah Roberts" |
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| Title: Re: const does not apply to pointers? |
30 Aug 2006 11:28:05 AM |
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Ron Natalie wrote:
Noah Roberts wrote:
It is for reasons like this that I have stopped using prefix const
notation. Even though it speaks better, "a constant int pointer," it
creates inconsistancies such as this. So now I say, "an int that is
contstant pointer."
What? "an int that is a pointer?"
Note that that is a rewording of my actual statement, one that is
iscompatible with the original.
Ints aren't pointers.
Thanks for the info.
How about a "pointer that is constant to int"
Yeah, that doesn't match the source code. The point I was making is
that "const int *" ends up reading like "constant int pointer." The
other way, "int const *," ends up reading, "int that is constant
pointer." Your statement would be equivelant to "* const int," and
that is a syntax error. It is also not equivelant to "const int *" as
it would be more akin to "int * const" or "an int pointer that is
constant."
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| User: "Ron Natalie" |
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| Title: Re: const does not apply to pointers? |
30 Aug 2006 01:48:13 PM |
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Noah Roberts wrote:
Yeah, that doesn't match the source code.
Once you get past a trivial declaration you're not going
to be able to linearly translate into something lexically
symentric. I say:
const char STAR foo
If I'm reading the code, but if I'm describing the
type I say
pointer to const char.
Trying to say something like
int *(*foo)(const char*, int(*)());
is just going to be gibberish read left to right.
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| User: "Noah Roberts" |
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| Title: Re: const does not apply to pointers? |
01 Sep 2006 10:32:51 AM |
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Ron Natalie wrote:
Noah Roberts wrote:
Yeah, that doesn't match the source code.
Once you get past a trivial declaration you're not going
to be able to linearly translate into something lexically
symentric. I say:
const char STAR foo
If I'm reading the code, but if I'm describing the
type I say
pointer to const char.
Trying to say something like
int *(*foo)(const char*, int(*)());
is just going to be gibberish read left to right.
Yeah, well do whatever you want. I find code more readable when it
reads clearly. I leave obscurity to those rare places where it's
needed. This also means I like sticking to "trivial" declarations
almost all the time.
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| User: "Stuart Redmann" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 03:40:21 AM |
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wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
This is one of the drawbacks of C++. Since the compiler can only ensure
that the binary representation of the object doesn't change, you may
change objects that are pointed to (since you don't change values of the
pointer, the 'binary' constness of your constant object is satisfied).
This logic is only expected, since it makes the compiler so much
simpler. You can avoid this pitfall by providing a proper interface to
your object (permitting access to a member variable is considered bad
style by the majority of C++ programmers):
#include <stdio.h>
class MyClass {
protected:
char *pChar; // Member variable is now protected.
public:
char *GetpChar () { // Accessor for pChar.
return pChar;
}
const char* GetpChar () const { // Another accessor for pChar
return pChar; // that gets invoked for const objects.
}
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void UsePointedTo (const char* ch)
{
printf (ch);
}
void DoStuff (const MyClass& constClass)
{
ChangePointedTo ( constClass.GetpChar ()); // this doesn't work anymore
}
void DoStuff2 (MyClass& non_constClass)
{
ChangePointedTo ( non_constClass.GetpChar ()); // this works
}
void DoStuff3 (const MyClass& constClass)
{
UsePointedTo ( constClass.GetpChar ()); // this works also
}
Regards,
Stuart
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| User: "" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 05:12:24 AM |
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I did a short skit on some of the vagaries of const for The Register
http://www.regdeveloper.co.uk/2006/07/26/constants_are_not/
DominiConnor
DCFC The Pimp
Stuart Redmann wrote:
grubertm@gmail.com wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
This is one of the drawbacks of C++. Since the compiler can only ensure
that the binary representation of the object doesn't change, you may
change objects that are pointed to (since you don't change values of the
pointer, the 'binary' constness of your constant object is satisfied).
This logic is only expected, since it makes the compiler so much
simpler. You can avoid this pitfall by providing a proper interface to
your object (permitting access to a member variable is considered bad
style by the majority of C++ programmers):
#include <stdio.h>
class MyClass {
protected:
char *pChar; // Member variable is now protected.
public:
char *GetpChar () { // Accessor for pChar.
return pChar;
}
const char* GetpChar () const { // Another accessor for pChar
return pChar; // that gets invoked for const objects.
}
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void UsePointedTo (const char* ch)
{
printf (ch);
}
void DoStuff (const MyClass& constClass)
{
ChangePointedTo ( constClass.GetpChar ()); // this doesn't work anymore
}
void DoStuff2 (MyClass& non_constClass)
{
ChangePointedTo ( non_constClass.GetpChar ()); // this works
}
void DoStuff3 (const MyClass& constClass)
{
UsePointedTo ( constClass.GetpChar ()); // this works also
}
Regards,
Stuart
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| User: "Old Wolf" |
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| Title: Re: const does not apply to pointers? |
31 Aug 2006 12:47:43 AM |
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wrote:
I did a short skit on some of the vagaries of const for The Register
http://www.regdeveloper.co.uk/2006/07/26/constants_are_not/
You call it a "skit", implying it is a joke. This is appropriate, since
it is almost completely factually incorrect. But if an average person
with only a modest familiarity with C++ were to read the article,
they could well take it seriously.
.
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| User: "Pete Becker" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 05:52:10 AM |
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Stuart Redmann wrote:
grubertm@gmail.com wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
This is one of the drawbacks of C++. Since the compiler can only ensure
that the binary representation of the object doesn't change, you may
change objects that are pointed to (since you don't change values of the
pointer, the 'binary' constness of your constant object is satisfied).
It's not a drawback. If you want to say that the thing that's pointed to
can't be changed through this pointer, mark it const:
const char *pChar;
.
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| User: "Earl Purple" |
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| Title: Re: const does not apply to pointers? |
29 Aug 2006 05:53:51 AM |
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Stuart Redmann wrote:
grubertm@gmail.com wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
<snip>
use string or vector<char> as the member instead of const char * and it
would work as desired.
There is no formal concept of ownership in C++ with regards to
pointers. If we talk about the state of a class changing, that should
also indicate anything it has ownership of, but not anything it simply
has reference of.
You cannot guarantee therefore that if T is a class and x() is a
const-method, then calling x() twice in succession will give the same
result even if there are no other threads because although the T won't
change, there may be other classes that change when you call x(). const
method doesn't mean change nothing.
Now I think Herb Sutter even addressed this issue at some point with
regards to propagating constness and there were proposals for different
levels of const. I think it was decided that whilst it might be useful
it made const-correctness even more difficult than it is now and would
make life too difficult for programmers to use.
The problem is more likely to sort itself out with the increasing use
of smart-pointers. It is a good idea in my opinion to implement those
smart-pointers that take ownership of their pointer (not shared_ptr
which only has shared ownership) to propagate constness. For smart
pointers that use a boolean flag to indicate ownership, I think this
would be impossible as it's a compiler error we are looking for, not a
runtime one. Of course this would probably lead whoever implements such
smart pointers to end up using two variations of templates with a
boolean-value template parameter (I've seen boost do that for traits
etc).
.
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| User: "Thomas Tutone" |
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| Title: Re: const does not apply to pointers? |
28 Aug 2006 09:07:49 PM |
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wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *" but instead it gets passed
to the function without any problems and suddenly MyClass is not so
const anymore...
You have confused a "const char*" with "char* const". Please take a
look at the FAQ:
http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.5
Best regards,
Tom
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| User: "Daniel T." |
|
| Title: Re: const does not apply to pointers? |
28 Aug 2006 09:31:59 PM |
|
|
In article <1156816029.862897.22730@75g2000cwc.googlegroups.com>,
wrote:
When accessing pointer member variables these are not treated as const
even though the instance is const. This came as a complete surprise to
me and I would appreciate some feedback why that is the case.
Here's a snippet:
class MyClass {
public:
char *pChar;
};
void ChangePointedTo (char *ch)
{
*ch = 0;
}
void DoStuff (const MyClass &constClass)
{
ChangePointedTo ( constClass.pChar ); // <-- why does this work?
}
I would have expected a compiler error telling me that
"constClass.pChar" is of type "const char *"
It's not "const char *", it's "char * const" which of course *can* be
passed to ChangePointerTo.
.
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