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Question on static and dynamic memory allocation

 
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KK
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PostPosted: Fri Nov 19, 2004 3:37 pm    Post subject: Question on static and dynamic memory allocation Reply with quote



Greetings,
I am trying to undersand the process of static and dynamic memory
allocation for a two dimensional array.

/* Example to illustrate the processes of memory
allocation of a two dimensional array */
#include<iostream>
using namespace std;
int main(){
/*Static memory allocation */
cout<<"Case : static array allocation:n";
char A[2][3];
cout<<"&A["<<0<<"] = "<<(void *)(&A[0])< cout<<"A[0] ="<<(void *)A[0]<
/*Dynamic memory allocation */
cout<<"Case : dynamic array allocation:n";
char **B=new char*[2];
for(int i = 0; i < 2; i++)
B[i]= new char[3];
cout<<"&B["<<0<<"] = "<<(void *)(&B[0])< cout<<"B[0] ="<<(void *)B[0]< return 0;
}

The output of the program is as follows,
Case : static array allocation:
&A[0] = 0012FF78
A[0] = 0012FF78

Case : dynamic array allocation:
&B[0] = 00320930
B[0] = 00320968

As you might have observed &A[0]==A[0] (static case) where as
&B[0]!=B[0](dynamic case). Could anyone kindly reason out this
inconsistent behaviour?
Thank you in advance,
-KK

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jjr2004a
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PostPosted: Tue Nov 23, 2004 10:45 am    Post subject: Re: Question on static and dynamic memory allocation Reply with quote



[email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote in message news:<c8fd5039.0411182132.1aab5b9f (AT) posting (DOT) google.com>...
Quote:
Greetings,
I am trying to undersand the process of static and dynamic memory
allocation for a two dimensional array.

/* Example to illustrate the processes of memory
allocation of a two dimensional array */
#include<iostream
using namespace std;
int main(){
/*Static memory allocation */
cout<<"Case : static array allocation:n";
char A[2][3];
cout<<"&A["<<0<<"] = "<<(void *)(&A[0])< cout<<"A[0] ="<<(void *)A[0]<
/*Dynamic memory allocation */
cout<<"Case : dynamic array allocation:n";
char **B=new char*[2];
for(int i = 0; i < 2; i++)
B[i]= new char[3];
cout<<"&B["<<0<<"] = "<<(void *)(&B[0])< cout<<"B[0] ="<<(void *)B[0]< return 0;
}

The output of the program is as follows,
Case : static array allocation:
&A[0] = 0012FF78
A[0] = 0012FF78

Case : dynamic array allocation:
&B[0] = 00320930
B[0] = 00320968

As you might have observed &A[0]==A[0] (static case) where as
&B[0]!=B[0](dynamic case). Could anyone kindly reason out this
inconsistent behaviour?
Thank you in advance,

The behavior is not inconsistent at all. By taking a look at
how the memory is laid out you can see why.

A is an array of six characters laid out in contiguous memory
like this:

+-+-+-+-+-+-+
Quote:
0|1|2|3|4|5|
+-+-+-+-+-+-+


B is slightly more complicated layout. It consists of space
for three pointers and space for six characters which are
not contiguous.

B B[0]

+-+ +-+ +-+-+-+
Quote:
*--->|*--->|0|1|2|
+-+ +-+ +-+-+-+ +-+-+-+
*------------>|3|4|5|
+-+ +-+-+-+


B[1]


From this layout you can deduce that B == &B[0], B[0] == &B[0][0]
and B[1] == &B[1][0] by definition.

This is really a C question. I've found the book "Advanced C Tips
and Techniques", by Kochan & Wood explains this stuff in depth.
It may be out of print but they seem to have a website and it may
be available from there for download.

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