Newbie question: Array memory address

[jose luis fernandez diaz]

Hi,

Given the program below:



#include <iostream>

int main()
{
char s1[2];
char *ptr = s1;

cout << sizeof(void *) << " " << sizeof(unsigned long) << endl;
cout << reinterpret_cast<unsigned long*>(s1) << endl;
cout << reinterpret_cast<unsigned long*>(&s1) << endl;
cout << reinterpret_cast<unsigned long*>(ptr) << endl;
cout << reinterpret_cast<unsigned long*>(&ptr) << endl;
}



PROGRAM OUTPUT:

frato2p:/tmp>a.out
8 8
0x11fffbff8
0x11fffbff8
0x11fffbff8
0x11fffbff0



memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?

Thanks,
Jose Luis.

 

[John Harrison]

Hi,

Given the program below:



#include <iostream>

int main()
{
char s1[2];
char *ptr = s1;

cout << sizeof(void *) << " " << sizeof(unsigned long) << endl;
cout << reinterpret_cast<unsigned long*>(s1) << endl;
cout << reinterpret_cast<unsigned long*>(&s1) << endl;
cout << reinterpret_cast<unsigned long*>(ptr) << endl;
cout << reinterpret_cast<unsigned long*>(&ptr) << endl;
}



PROGRAM OUTPUT:

frato2p:/tmp>a.out
8 8
0x11fffbff8
0x11fffbff8
0x11fffbff8
0x11fffbff0



memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?

Thanks,
Jose Luis.

It says the address of an array and the address of its first element are the
same.

But the address of a pointer and the address it points to are not
(necessarily) the same.

So your output is correct.

john

 

[Ron Natalie]

int main()
{
char s1[2];

memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?

Why does this surprise you. In the case of memory_address(s1) you convert
an array to a pointer to it's first element. In the case of memory_address(&s1)
you are taking the address of the array. It's not surprising that the address
of the array and it's first element are the same.

You do understand the difference between an array and a pointer?

 

[John]

int main()
{
char s1[2];

memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?

Why does this surprise you. In the case of memory_address(s1) you convert
an array to a pointer to it's first element. In the case of memory_address(&s1)
you are taking the address of the array. It's not surprising that the address
of the array and it's first element are the same.

You do understand the difference between an array and a pointer?

s1 is a pointer, right? Is &s1 the address to pointer s1?

john

 

[Jack Klein]

int main()
{
char s1[2];

memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?

Why does this surprise you. In the case of memory_address(s1) you convert
an array to a pointer to it's first element. In the case of memory_address(&s1)
you are taking the address of the array. It's not surprising that the address
of the array and it's first element are the same.

You do understand the difference between an array and a pointer?

s1 is a pointer, right? Is &s1 the address to pointer s1?

Wrong, s1 is not a pointer. An array is not a pointer, a pointer is
not an array. If you do not learn that well, you will have constant
and enormous difficulties in C++ and C.

The NAME of an array, in many contexts, is converted to a pointer to
the first element of the array. The primary exceptions are when you
apply the sizeof operator to the name of an array, and when you apply
the & operator to the name of an array.

 

[jose luis fernandez diaz]

By tbe way, how can you get an array variable address ?

Thanks,
Jose Luis.

jose luis fernandez diaz wrote:

int main()
{
char s1[2];
memory_address(s1) == memory_address(&s1). Why does the Standard say about this ?


Why does this surprise you. In the case of memory_address(s1) you convert
an array to a pointer to it's first element. In the case of memory_address(&s1)
you are taking the address of the array. It's not surprising that the address
of the array and it's first element are the same.

You do understand the difference between an array and a pointer?

s1 is a pointer, right? Is &s1 the address to pointer s1?

Wrong, s1 is not a pointer. An array is not a pointer, a pointer is
not an array. If you do not learn that well, you will have constant
and enormous difficulties in C++ and C.

The NAME of an array, in many contexts, is converted to a pointer to
the first element of the array. The primary exceptions are when you
apply the sizeof operator to the name of an array, and when you apply
the & operator to the name of an array.

 

[Ron Natalie]

By tbe way, how can you get an array variable address ?

You put & in front of the name. I have a feeling that you still
don't understand what an array is. There is no "variable" other
than the array itself.

When you declare
int a[10];

it makes 10 ints in memory. &a is the beginning of the 10 ints.
When you use the array value, it converts to a pointer to the first
int in the array, whose type is different, but refers to the same
location in memory (the representation of the pontier may also
be different as some computers have different flavors of pointers
for different object types).

 

[jose luis fernandez diaz]

#include <iostream>
#include <limits>
#include <set>

int main()
{
int i=1;
int *ptr_i=&i;
int **ptr_ptr_i=&ptr_i;

cout << reinterpret_cast<unsigned long*>(&i) << endl;
cout << reinterpret_cast<unsigned long*>(ptr_i) << endl;
cout << *ptr_i << endl;
cout << **ptr_ptr_i << endl;
}

frato2p:/tmp>a.out
0x11fffbff0
0x11fffbff0
1
1

When you apply the '&' operator to a variable, you get the mermory
address where it resides. In the program above, 'ptr_i' holds the
memory address where 'i' resides. How can I get the memory address
where an array varible resides ?

Thanks,
Jose Luis

By tbe way, how can you get an array variable address ?

You put & in front of the name. I have a feeling that you still
don't understand what an array is. There is no "variable" other
than the array itself.

When you declare
int a[10];

it makes 10 ints in memory. &a is the beginning of the 10 ints.
When you use the array value, it converts to a pointer to the first
int in the array, whose type is different, but refers to the same
location in memory (the representation of the pontier may also
be different as some computers have different flavors of pointers
for different object types).

 

[John Harrison]

#include <iostream>
#include <limits>
#include <set>

int main()
{
int i=1;
int *ptr_i=&i;
int **ptr_ptr_i=&ptr_i;

cout << reinterpret_cast<unsigned long*>(&i) << endl;
cout << reinterpret_cast<unsigned long*>(ptr_i) << endl;
cout << *ptr_i << endl;
cout << **ptr_ptr_i << endl;
}

frato2p:/tmp>a.out
0x11fffbff0
0x11fffbff0
1
1

When you apply the '&' operator to a variable, you get the mermory
address where it resides. In the program above, 'ptr_i' holds the
memory address where 'i' resides. How can I get the memory address
where an array varible resides ?

Thanks,
Jose Luis

int a[10];
int* ap1 = a;
int (*ap2)[10] = &a;
cout << ap1 << '\n';
cout << ap2 << '\n';

Either &a or a is the address where the array resides.

john

 

[Ron Natalie]

When you apply the '&' operator to a variable, you get the mermory
address where it resides. In the program above, 'ptr_i' holds the
memory address where 'i' resides. How can I get the memory address
where an array varible resides ?

The array variable is the array.
int foo[10];
&foo IS where the array resides. It's also the same location
as (int*) foo, which is the first element of the array. There
is NOTHING separate abou the array foo other than the ten ints
that comprise it.

ptr_i in your example is a seperate boject from what it points to.


If you want to declare a pointer to the array, declare it as
int (*ptr_foo)[10] = &foo;

 

[jose luis fernandez diaz]

OK. I understand.

Best regards,
Jose Luis.

When you apply the '&' operator to a variable, you get the mermory
address where it resides. In the program above, 'ptr_i' holds the
memory address where 'i' resides. How can I get the memory address
where an array varible resides ?

The array variable is the array.
int foo[10];
&foo IS where the array resides. It's also the same location
as (int*) foo, which is the first element of the array. There
is NOTHING separate abou the array foo other than the ten ints
that comprise it.

ptr_i in your example is a seperate boject from what it points to.


If you want to declare a pointer to the array, declare it as
int (*ptr_foo)[10] = &foo;