I have some codes:
w_char **sFieldsUri;
sFieldsUri = &(new w_char[2]);

w_char **sValFind = nsnull;
sValFind = &(new w_char[2]);

When I debug my program, I found the above code had some strange
things. sFieldsUri+1 is equal to sValFind+0.

But if I change the codes to the following:
w_char **sFieldsUri;
sFieldsUri = &(new w_char[2]);

for(int i=0; i<2; i++)
{
sFieldsUri[i] = new w_char[100];
}

w_char **sValFind = nsnull;
sValFind = &(new w_char[2]);

for(int i=0; i<2; i++)
{
sVal Find[i] = new w_char[100];
}
sFieldsUri+0, sFieldsUri+1, and sValFind+0, sValFind+1 are all
difference.
Thanks for all.

DDD said:

> I have some codes:
> w_char **sFieldsUri;
> sFieldsUri = &(new w_char[2]);

The language rules for C and C++ differ. I suggest you ask this C++
question in comp.lang.c++ for best results.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

DDD wrote:
> I have some codes:
> w_char **sFieldsUri;
> sFieldsUri = &(new w_char[2]);
^^^^^^^^^
This tells us that you are in the wrong place, since that is a syntax
error in C. Try posting to a newsgroup for whatever language you are using.

DDD <1983ddd@gmail.com> wrote:

> I have some codes:
> w_char **sFieldsUri;
> sFieldsUri = &(new w_char[2]);

This is not C, but (probably!) C++. Pointer handling in C and C++ are
sufficiently different that you should ask this in comp.lang.c++,
because any answer you get here will be given from a C POV, and might
therefore not apply to C++.

Richard

On Jan 28, 2:08 am, DDD <1983...@gmail.com> wrote:
> I have some codes:
> w_char **sFieldsUri;
> sFieldsUri = &(new w_char[2]);

new w_char[2] will return a w_char*, so far so good. Except you are
trying to take the address of non lvalue ...
Enable ALL warning/error messages from your compiler, you should get
an error.
&(new w_char[2]) is really weird, really (besides from taking the
adress of non lvalue) it is like wrting int* p = &3;

>
> w_char **sValFind = nsnull;
> sValFind = &(new w_char[2]);
>
> When I debug my program, I found the above code had some strange
> things. sFieldsUri+1 is equal to sValFind+0.

I am still surprised that your compiler can compile this.

>
> But if I change the codes to the following:
> w_char **sFieldsUri;
> sFieldsUri = &(new w_char[2]);

forget it.

>
> for(int i=0; i<2; i++)
> {
> sFieldsUri[i] = new w_char[100];
> }
>
> w_char **sValFind = nsnull;
> sValFind = &(new w_char[2]);
>
> for(int i=0; i<2; i++)
> {
> sVal Find[i] = new w_char[100];
> }
> sFieldsUri+0, sFieldsUri+1, and sValFind+0, sValFind+1 are all
> difference.
> Thanks for all.

If I were a gambler I would say you want that:

w_char **sFieldsUri;
sFieldsUri = new w_char*[2]; // notice the '*' thing

and try to re-run/compile your program.

you should get a decent compiler dude.

Cheers,
Paulo

ppi wrote:
> On Jan 28, 2:08 am, DDD <1983...@gmail.com> wrote:
>> I have some codes:
>> w_char **sFieldsUri;
>> sFieldsUri = &(new w_char[2]);
>
> new w_char[2] will return a w_char*,

Not it C, it won't. It's a syntax error.
so far so good.

Bullshit.