«

About 20 days ago, "John Harrison" <>
replied to a message which I had written:

> > > Q1 . What is the use of pointer to an array?
> >
> > It points to it. (The name of an array is also a pointer to it.)
>
> No, the name of an array decays to a pointer to the first element of the
> array. A pointer to an array and a pointer to the first element of an array
> are not the same thing.
>
> int a[10];
> cout << &a + 1 << endl;
> cout << a + 1 << endl;
>
> The two statements print different pointer values, proving that a pointer to
> an array (first case) and a pointer to the first element of an array (second
> case) are not the same.

I was dubious, so I wrote this program:

#include <iostream>
using std::cout;
using std::endl;
int main(void)
{
int a[10];
cout << "int a[10];" << endl;
cout << " &a : " << (&a) << endl;
cout << " a : " << ( a) << endl;
cout << "&a + 1 : " << (&a + 1) << endl;
cout << " a + 1 : " << ( a + 1) << endl;
return 0;
}

and I got these results:

wd=C:\C\test
%array-pointer-test
int a[10];
&a : 0x6cffac
a : 0x6cffac
&a + 1 : 0x6cffd4
a + 1 : 0x6cffb0

Interestingly, the value of &a and a are the same. But when incremented,
one increments 4 bytes, the other 40 bytes.

Of course. They're pointers to different types, one of size 4 and the other
of size 40. I should have known that. ::: kicks self :::

In C, however, depending on the context, the compiler is likely to
implicitly cast the pointer from pointer-to-array to pointer-to-int.
For example, consider the following code:

/* char-array-test.c */
#include <stdio.h>
char Func(char* blat)
{
return *(blat+1);
}
int main(void)
{
char a[80];
a[0] = 'n';
a[1] = 't';
printf("Func(&a) = %c\n", Func(&a));
return 88;
}

This prints 't' in C, but it won't even compile in C++.

I think I shouldn't have learned C before C++; I've picked up some bad
habits that way.

--
Cheers,
Robbie Hatley
Tustin, CA, USA
email: lonewolfintj at pacbell dot net
web: home dot pacbell dot net slant earnur slant







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"Robbie Hatley" <lonewolfintj at pacbell dot net> wrote...
> [...]
> I was dubious, so I wrote this program:
>
> #include <iostream>
> using std::cout;
> using std::endl;
> int main(void)
> {
> int a[10];
> cout << "int a[10];" << endl;
> cout << " &a : " << (&a) << endl;
> cout << " a : " << ( a) << endl;
> cout << "&a + 1 : " << (&a + 1) << endl;
> cout << " a + 1 : " << ( a + 1) << endl;
> return 0;
> }
>
> and I got these results:
>
> wd=C:\C\test
> %array-pointer-test
> int a[10];
> &a : 0x6cffac
> a : 0x6cffac
> &a + 1 : 0x6cffd4
> a + 1 : 0x6cffb0
>
> Interestingly, the value of &a and a are the same. But when incremented,
> one increments 4 bytes, the other 40 bytes.
>
> Of course. They're pointers to different types, one of size 4 and the
other
> of size 40. I should have known that. ::: kicks self :::
>
> In C, however, depending on the context, the compiler is likely to
> implicitly cast the pointer from pointer-to-array to pointer-to-int.

I just wonder how you come to this conclusion.

> For example, consider the following code:
>
> /* char-array-test.c */
> #include <stdio.h>
> char Func(char* blat)
> {
> return *(blat+1);
> }
> int main(void)
> {
> char a[80];
> a[0] = 'n';
> a[1] = 't';
> printf("Func(&a) = %c\n", Func(&a));

So, inside that function the address that you obtain by &a, is still
interpreted as char*, as if you passed 'a' (instead of &a). What is
surprising about that?

In C, anything used to be convertible to char* (and back). Similar to
void* in C++ (and now in C99, IIRC). The difference is that you can't
really do anything with void*, whereas char* can be interpreted as
a string (or array of characters). IMO, C++ fixed what was kind of
broken in C. Or at least tried to fix it. C gived char* dual meaning
-- a generic address holder and a pointer to [an array of] character[s].

> return 88;
> }
>
> This prints 't' in C, but it won't even compile in C++.

Of course it won't compile in C++. VC++ 7.1 (when compiling C) does
actually warn about the difference in "levels of indirection". It
lets the difference slide simply because C was built to serve that.

> I think I shouldn't have learned C before C++; I've picked up some bad
> habits that way.

Learing C _correctly_ before C++ can't hurt, at least not too much.
Bad habits can be picked in either language and without them. Try
not to blame languages for the way they are [mis]used.

Also, knowing the differnces between the languages and how they do
things they do and what is allowed and not allowed there always
helps.

Don't beat yourself too much...

Victor

On Fri, 4 Jun 2004 20:04:53 -0700, "Robbie Hatley" <lonewolfintj at
pacbell dot net> wrote in comp.lang.c++:

> About 20 days ago, "John Harrison" <>
> replied to a message which I had written:
>
> > > > Q1 . What is the use of pointer to an array?
> > >
> > > It points to it. (The name of an array is also a pointer to it.)
> >
> > No, the name of an array decays to a pointer to the first element of the
> > array. A pointer to an array and a pointer to the first element of an array
> > are not the same thing.
> >
> > int a[10];
> > cout << &a + 1 << endl;
> > cout << a + 1 << endl;
> >
> > The two statements print different pointer values, proving that a pointer to
> > an array (first case) and a pointer to the first element of an array (second
> > case) are not the same.
>
> I was dubious, so I wrote this program:
>
> #include <iostream>
> using std::cout;
> using std::endl;
> int main(void)
> {
> int a[10];
> cout << "int a[10];" << endl;
> cout << " &a : " << (&a) << endl;
> cout << " a : " << ( a) << endl;
> cout << "&a + 1 : " << (&a + 1) << endl;
> cout << " a + 1 : " << ( a + 1) << endl;
> return 0;
> }
>
> and I got these results:
>
> wd=C:\C\test
> %array-pointer-test
> int a[10];
> &a : 0x6cffac
> a : 0x6cffac
> &a + 1 : 0x6cffd4
> a + 1 : 0x6cffb0
>
> Interestingly, the value of &a and a are the same. But when incremented,
> one increments 4 bytes, the other 40 bytes.
>
> Of course. They're pointers to different types, one of size 4 and the other
> of size 40. I should have known that. ::: kicks self :::
>
> In C, however, depending on the context, the compiler is likely to
> implicitly cast the pointer from pointer-to-array to pointer-to-int.
> For example, consider the following code:

Actually, depending on the compiler ***and the way you invoke it***.

If you invoke your compiler as a conforming C compiler (which none I
know of are by default), it must issue a diagnostic when you pass a
pointer to an array to a function prototyped as accepting a pointer to
char.

Asking the compiler to perform that conversion explicitly is a
constraint violation.

Check the compiler's documentation and invoke it in standard C
conforming mode, and it will no more accept the program without
complaint than will a C++ compiler.

--
Jack Klein
Home: http://JK-Technology.Com
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