pass by Reference/value ???

[dumboo]

hi there
i m little bit confused over the following problem, i have understood wt the
following code is doing...but not able to get the actual techinical
stuff.....i have had a lot of hot debate over the following code whether its
pass by reference or pass by value...hope somebody clears it up :-)
void foo(char *s)
{
s = new char[10];
strcpy(s, "gotit");
}

int main()
{
char *s;
foo(s);
cout<<s;
delete s;

return 0;
}

regards

[Default User]

Jonathan Turkanis wrote:

"dumboo" <vt***@yahoo.co m> wrote in message
news:c0******** *****@ID-211285.news.uni-berlin.de...

hi there
i m little bit confused over the following problem, i have

understood wt the

following code is doing...but not able to get the actual techinical
stuff.....i have had a lot of hot debate over the following code

whether its

pass by reference or pass by value...hope somebody clears it up :-)
void foo(char *s)
{
s = new char[10];
strcpy(s, "gotit");
}

int main()
{
char *s;
foo(s);
cout<<s;
delete s;

return 0;
}

Here you are passing a pointer by value.

Among other things, you need to delete s like so: delete [] s.

That just somewhat different undefined behavior to the undefined
behavior he already has due to his faulty allocation.

The pointer back in main is unaffected by the new over in foo(). So
deferencing it in main() via the cout is undefined. The delete is wrong,
but so is using delete[] on an uninitialized pointer.

The function foo() either needs to take a char** or return a char*.

Brian Rodenborn

[Default User]

John Carson wrote:

If you want to modify s, then there are two ways to go about it. The C-style
approach is to pass the address of s rather than s itself, as follows:

void foo(char **ps)
{
*ps = new char[10];
strcpy(*ps, "gotit");
} The C++-style approach is to use references as follows:

void foo(char *& s)
{
s = new char[10];
strcpy(s, "gotit");
}

A third way is to return a pointer, like so:

char* foo(char *s)
{
s = new char[10];
strcpy(s, "gotit");
return s;
}
Brian Rodenborn

[Old Wolf]

> > hi there

i m little bit confused over the following problem, i have understood wt the
following code is doing...but not able to get the actual techinical
stuff.....i have had a lot of hot debate over the following code whether its
pass by reference or pass by value...hope somebody clears it up :-)

int main()
{
char *s;

`s' is an uninitialized pointer-to-char that points to...well, *anywhere*,

foo(s);

`foo()' is called; however as it has been passed by value it is
unchanged, hence still uninitialized.

cout<<s;

An attempt is made to send the contents pointed to by `s' to the stream
`cout'. Since `s' points...well *anywhere*...un defined behavior is
invoked. [first time]

Second time, actually; foo(s) is UB because it evaluates s
(it could be a trap value).

[John Carson]

"Default User" <fi********@boe ing.com.invalid > wrote in message
news:40******** *******@boeing. com.invalid

A third way is to return a pointer, like so:

char* foo(char *s)
{
s = new char[10];
strcpy(s, "gotit");
return s;
}

In which case no use is being made of the argument passed except as a source
of a local variable, so it would be simpler to use:

char* foo()
{
char *s = new char[10];
strcpy(s, "gotit");
return s;
}
--
John Carson
1. To reply to email address, remove donald
2. Don't reply to email address (post here instead)

[Sharad Kala]

"Default User" <fi********@boe ing.com.invalid > wrote in message
news:40******** *******@boeing. com.invalid...

John Carson wrote:

If you want to modify s, then there are two ways to go about it. The C-style
approach is to pass the address of s rather than s itself, as follows:

void foo(char **ps)
{
*ps = new char[10];
strcpy(*ps, "gotit");
}

The C++-style approach is to use references as follows:

void foo(char *& s)
{
s = new char[10];
strcpy(s, "gotit");
}

A third way is to return a pointer, like so:

char* foo(char *s)
{
s = new char[10];
strcpy(s, "gotit");
return s;
}

For a simple program like this, it's fine.
But atleast in a library I would not like this approach.
This would mean that you need to tell your users that the responsibility to free
the memory is with you.
Unless one is quite careful with such libraries there are bound to be potential
memory leaks.

-Sharad