About c++ pointer

[zhou xiang]

/////////////// code 1 start ///////////////////////
char *GetString(void )
{
char p[] = "hello world";
return p;
}
void Test4(void)
{
char *str = NULL;
str = GetString();
cout<< str << endl;
}
/////////////// code 1 end ////////////////////////

///////////////// code 2 start ////////////////////
char *GetString2(voi d)
{
char *p = "hello world";
return p;
}
void Test5(void)
{
char *str = NULL;
str = GetString2();
cout<< str << endl;
}
///////////////// code 2 end /////////////////////

Q: What is the difference between code 1 and code 2?

[Andrey]

zhou xiang wrote:

/////////////// code 1 start ///////////////////////
char *GetString(void )
{
char p[] = "hello world";
return p;
}
void Test4(void)
{
char *str = NULL;
str = GetString();
cout<< str << endl;
}
/////////////// code 1 end ////////////////////////

///////////////// code 2 start ////////////////////
char *GetString2(voi d)
{
char *p = "hello world";
return p;
}
void Test5(void)
{
char *str = NULL;
str = GetString2();
cout<< str << endl;
}
///////////////// code 2 end /////////////////////

Q: What is the difference between code 1 and code 2?

The code 1 will not work, the pointer will be empty when it returns,
because the object(string array) is created in stack. While the second
is fine since you dynamically allocate memory in the heap.

[Peter Koch Larsen]

"zhou xiang" <ju********@eyo u.com> skrev i en meddelelse
news:82******** *************** ***@posting.goo gle.com...

/////////////// code 1 start ///////////////////////
char *GetString(void )
{
char p[] = "hello world";
Here you declare a local array and initialise it to "Hello world".
return p; .... and you return a pointer to that array, and deallocate it - thus p will
likely point to garbage. }
void Test4(void)
{
char *str = NULL;
str = GetString();
cout<< str << endl;
This is undefined behaviour and might print garbage or stop the program.
}
/////////////// code 1 end ////////////////////////

///////////////// code 2 start ////////////////////
char *GetString2(voi d)
{
char *p = "hello world";
Here you declare a pointer which points to a constant static area containing
"hello world". This is okay, but dangerous: the area could be changed (e.g.
by executing p[0] = 'H'), but that would lead to undefined behaviour.
The above line should read
char const* p= ...;
return p;
}
void Test5(void)
{
char *str = NULL;
str = GetString2();
cout<< str << endl;
}

/Peter

[Matthias Käppler]

Andrey wrote:

the second is fine since you dynamically allocate memory in the heap.

In how far is 'char * p = "hallo";' dynamic memory allocation?
The pointer p will point to nothing as well when leaving the scope.

Am I missing something?

Regards,
Matthias

[abcd]

Andrey <st*******@yaho o.ca> wrote in message news:<rzgod.296 877$Pl.79890@pd 7tw1no>...

zhou xiang wrote:
[...]

Q: What is the difference between code 1 and code 2?

The code 1 will not work, the pointer will be empty when it returns,
because the object(string array) is created in stack. While the second
is fine since you dynamically allocate memory in the heap.

Nope. In neither case is the memory allocated off the heap (I can't
see any 'new' or 'malloc'). In both cases the string is statically
allocated in read-only memory (try to change it and the program will
likely crash e.g., *p='f').

The difference between the two is that in the first listing p is an
actual array, allocated on the stack. The contents of the constant
string is copied into it. It will be destroyed as soon as the flow of
control leaves function GetString(). This means that you're returning
the address of a local (and a clever compiler issues a warning). The
second listing is fine, not because you're allocating memory off the
heap (which you're not) but because you're returning a pointer to a
statically allocated string which will be there as long as the program
runs.

HTH
Andy

[Andrey]

Matthias Käppler wrote:

Andrey wrote:

the second is fine since you dynamically allocate memory in the heap.

In how far is 'char * p = "hallo";' dynamic memory allocation?
The pointer p will point to nothing as well when leaving the scope.

Am I missing something?

Regards,
Matthias

I think you are right. no memory allocation was done. It doesn't work
even in the scope of the function.
I thougt there was new, I missed that. mistake.
Regards, Andrey

[Andrey]

Andrey wrote:

Matthias Käppler wrote:

Andrey wrote:

the second is fine since you dynamically allocate memory in the heap.

In how far is 'char * p = "hallo";' dynamic memory allocation?
The pointer p will point to nothing as well when leaving the scope.

Am I missing something?

Regards,
Matthias

I think actually it would be good (in some sence) if you could write
like this:) But you always have to mess with char arrays allocating and
freeing memory for them. Ceratinly you should know how to do it but
that's annoying. That's why I like CString from MFC library, well
written class(template) , though I'm not thinking too well about the rest
of MFC. It'd be good to have an implementation of CString away from MFC
library. There're some alternatives certainly, but are there which are
as good as CString?

Andrey.

[Ali Çehreli]

Original code once again:

char *GetString2(voi d)
{
char *p = "hello world";
return p;
}
"Matthias Käppler" <no****@digital raid.com> wrote in message
news:cn******** *****@news.t-online.com...

Andrey wrote:

the second is fine since you dynamically allocate memory in the heap.
In how far is 'char * p = "hallo";' dynamic memory allocation?

You are right, no dynamic memory allocation above.
The pointer p will point to nothing as well when leaving the scope.

This is not correct though. The string literal "hello world" has static
duration and will be valid for the lifetime of the program.

p above points to that static storage, GetString2 returns a copy of p, still
pointing to that static storage. The callers of GetString2 can access the
returned pointer.

There is another point to be made though: The code better be written to use
'char const *' types. Because the string literal is actually constant:

char const * GetString2(void )
{
char const * p = "hello world";
return p;
}

The previous definition would make some compilers compile the code but cause
undefined behavior at runtime if in fact some caller tries to modify the
content:

char * returned = GetString2();
returned[0] = 'a'; // undefined behavior

If the function were written with 'char const *', the compiler would not
allow this code.

Ali

[Matthias Käppler]

Andrey wrote:

I think actually it would be good (in some sence) if you could write
like this:) But you always have to mess with char arrays allocating and
freeing memory for them. Ceratinly you should know how to do it but
that's annoying. That's why I like CString from MFC library, well
written class(template) , though I'm not thinking too well about the rest
of MFC. It'd be good to have an implementation of CString away from MFC
library. There're some alternatives certainly, but are there which are
as good as CString?

Andrey.

std::string? ^^

It's the string class from the C++ standard library. It doesn't have a
printf-like operation like CString's Format(), but apart from that it has
all the desired operations like operator+= for concatenation and so on.

I almost -never- use C-style strings. Too error prone. Too unhandy.

Regards,
Matthias

[Jonathan Mcdougall]

> /////////////// code 1 start ///////////////////////

char *GetString(void )
void in the parameter list is redundant and considered bad style.
{
char p[] = "hello world";
That's a local variable which will be destroyed...
return p;
....here. Therefore, the address returned is invalid.
}
void Test4(void)
{
char *str = NULL;
str = GetString();
str points nowhere.
cout<< str << endl;
oops.
}
/////////////// code 1 end ////////////////////////

///////////////// code 2 start ////////////////////
char *GetString2(voi d)
idem.
{
char *p = "hello world";

That should be

const char *p = "hello world";

This is different. The string is place somewhere in outer-space where
all things are const. IIRC, the string is guaranteed to exist for the
whole program. What I am sure of is that the string cannot be modified
(and your program will crash if you try to). Therefore, if you don't
touch that string, the program will run as expected.
Jonathan