C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Question on memory allocation process of a 2D array          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message KK Guest Posted: Thu Nov 18, 2004 12:49 am    Post subject: Question on memory allocation process of a 2D array Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. /* 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 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top James Hopkin Guest Posted: Fri Nov 19, 2004 12:53 pm    Post subject: Re: Question on memory allocation process of a 2D array [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote in message news:<c8fd5039.0411171317.2b51f5f8 (AT) posting (DOT) google.com>... Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. The two are not equivalent, the way you have written it. The static part of your example defines a genuine two dimensional array (in your case, six contiguous elements). The dynamic part allocates a one dimensional array of pointers. (Kernighan and Ritchie "The C Programming Language", 5.9, has a good illustration of the difference). The following allocates an equivalent dynamic array: char (*B)[3] = new char[2][3]; Note, however that B is a pointer to the first array in an array of arrays, whereas A *is* an array of arrays. In other words, &A[0] corresponds to B; A[0] corresponds to *B. James [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Richter Guest Posted: Fri Nov 19, 2004 12:53 pm    Post subject: Re: Question on memory allocation process of a 2D array Hi, Quote: From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. Sure, but your types are different, so why do you expect the same behaivour? Quote: /* Example to illustrate the processes of memory allocation of a two dimensional array */ #include using namespace std; int main(){ /*Static memory allocation */ cout<<"Case : static array allocation:n"; char A[2][3]; Here "A" is a two-dimensional array. Quote: 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]; Here "B" is a pointer to a one-dimensional array of pointers. Note that this is a *different* type than that of the A object. There is just an implicit conversion that converts "A" objects to pointer objects so that dereferencing them with [] works as expected. Quote: for(int i = 0; i < 2; i++) B[i]= new char[3]; cout<<"&B["<<0<<"] = "<<(void *)(&B[0])< Above, you're printing the address of the A array and the address of its first element. This is indeed the same as the array starts with its first element in the memory allocated for it. For the B type, you're printing the address of the pointer that points to the array of pointers, i.e. &B is a char ***. Quote: cout<<"B[0] ="<<(void *)B[0]< Here's a bug in your program: You should release all memory you've allocated. Even if you're just testing, it's good practise to start with a high disipline. Quote: return 0; } 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? Well, you're doing different things, so you should expect different things; A and B are different types, and completely differently layed out arrays. While A is a plain and simple array, B is a pointer to an array of pointers each pointing to an array. Thus, different results due to different layouts. What *is* confusing is that you are able to use both types in the same way, i.e. with [][]. However, even though these operations look the same, they do something completely different. So long, Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Steven Kilby Guest Posted: Fri Nov 19, 2004 12:59 pm    Post subject: Re: Question on memory allocation process of a 2D array Quote: /*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]< This is contiguous. Quote: /*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]< This is not. The variable B has a value that is the address of the memory allocated with new char*[2]. &B[0] is that same address. B[0] has a value that is the address of the memory allocated with new char[3]. This is a different value than the address OF B[0] which was allocated with the new char*[2]. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Branimir Maksimovic Guest Posted: Fri Nov 19, 2004 2:54 pm    Post subject: Re: Question on memory allocation process of a 2D array [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote in message news:<c8fd5039.0411171317.2b51f5f8 (AT) posting (DOT) google.com>... Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. ........ /*Dynamic memory allocation */ cout<<"Case : dynamic array allocation:n"; char **B=new char*[2]; not the same thing; you are allocating array of pointers in this case Quote: 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; you should do this: char(*B)[2][3] = (char(*)[2][3])new char[6]; cout<<"&B["<<0<<"] = "<<(void*)(&(*B)[0])< cout<<"B["<<0<<"] = "<<(void*)(*B)[0]< Greetings, Bane. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Tom Widmer Guest Posted: Fri Nov 19, 2004 3:03 pm    Post subject: Re: Question on memory allocation process of a 2D array On 17 Nov 2004 19:49:17 -0500, [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote: Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. /* Example to illustrate the processes of memory allocation of a two dimensional array */ #include 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]< If you want the dynamic allocation to be "equivalent", you should do: cout<<"Case : dynamic array allocation:n"; typedef char array[2]; char (*B)[2]= new array[3]; cout<<"&B["<<0<<"] = "<<(void *)(&B[0])< cout<<"B[0] ="<<(void *)B[0]< Quote: 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? The inconsistency is in your code - your auto case had an array of 3 arrays of 2 chars, whereas your dynamic case had a dynamic array of 3 dynamic arrays of char, each of size 2. With the above change, you end up with a dynamic array of 3 arrays of 2 chars, which is roughly equivalent to the auto case. Tom [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Chris Uzdavinis Guest Posted: Fri Nov 19, 2004 3:22 pm    Post subject: Re: Question on memory allocation process of a 2D array [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote in message news:<c8fd5039.0411171317.2b51f5f8 (AT) posting (DOT) google.com>... Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. /* Example to illustrate the processes of memory allocation of a two dimensional array */ #include 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]< Declaring A here, it is a real array. The memory for all the array is contiguous. Even though it's a two dimensional array, accessing A[0] returns to you an array of 3 characters. That array happens to begin at the same address, since this is all allocated directly on the stack. Quote: /*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; } Here, B is *not* an array. It is a pointer, and the value you made it point to is the address returned by your first call to new[]. You are comparing the address of a pointer against the address stored in that pointer. -- Chris [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Steven Kilby Guest Posted: Fri Nov 19, 2004 3:26 pm    Post subject: Re: Question on memory allocation process of a 2D array Quote: /*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]< This is contiguous. Quote: /*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]< This is not. The variable B has a value that is the address of the memory allocated with new char*[2]. &B[0] is that same address. B[0] has a value that is the address of the memory allocated with new char[3]. This is a different value than the address OF B[0] which was allocated with the new char*[2]. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Steven Kilby Guest Posted: Fri Nov 19, 2004 3:27 pm    Post subject: Re: Question on memory allocation process of a 2D array Quote: /*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]< This is contiguous. Quote: /*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]< This is not. The variable B has a value that is the address of the memory allocated with new char*[2]. &B[0] is that same address. B[0] has a value that is the address of the memory allocated with new char[3]. This is a different value than the address OF B[0] which was allocated with the new char*[2]. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Allan W Guest Posted: Fri Nov 19, 2004 4:08 pm    Post subject: Re: Question on memory allocation process of a 2D array [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote Quote: I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. C++ doesn't directly support two-dimensional arrays, but it does support an array of arrays, which has much the same effect. Quote: Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. /* Example to illustrate the processes of memory allocation of a two dimensional array */ #include 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]< A is an array of 2 arrays of 3 characters each. Quote: Case : static array allocation: &A[0] = 0012FF78 A[0] = 0012FF78 Quote: /*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; } But this is different. Here you're allocating an array of 2 character pointers and two arrays of 3 characters. (The delete code, which you don't show, will also require you to do a delete for each NEW). Quote: Case : dynamic array allocation: &B[0] = 00320930 B[0] = 00320968 &B[0] gives you the address of the first pointer. B[0] gives you the value of the first pointer. To do the same as the first approach, you would have to write: /* Dynamic allocation of a 2-by-3 array */ char (*C)[2][3] = new char[][2][3]; cout<<"&C[0] = "<<(void *)(&C[0])< cout<<" C[0] = "<<(void *)C[0]< delete[] C; &C[0] = 00322C30 C[0] = 00322C30 Better? [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Antoun Kanawati Guest Posted: Fri Nov 19, 2004 4:49 pm    Post subject: Re: Question on memory allocation process of a 2D array [email]kewlkarun (AT) yahoo (DOT) com[/email] (KK) wrote in message news:<c8fd5039.0411171317.2b51f5f8 (AT) posting (DOT) google.com>... Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. Actually, there are three cases: dynamic, automatic, and static. And this is pretty much for anything involving 'allocation'. As for 'process', it depends on what you mean by the word; and how to interpret 'exactly the same'. In C++, 2D array are stored in row-major order, in a contiguous chunk of memory. That's about all the sameness that I can think of. To allocate a static or automatic 2D array, you say something like: char a[4][5]; The 'static' process is to reserve the space at compile time, and merely point the symbol to it. You can expect the 20 chars to be zeroed out. The 'automatic' process involves a runtime adjustment of the stack pointer. The 20 chars will hold random junk in this case. To do a dynamic allocation: typedef char char5[5]; char5 *a = new char5[4]; // or char5 *a = new char[4][5]; In a well-behaved program, a corresponding delete must occur in the code. In all cases, we're allocating 20 chars, and interpreting the chunk as four rows of 5-chars each. In the dynamic case, you can avoid the typedef by doing some awkward parenthesization in the next line. However, as you can see, the "process" of creating the 2D array is not exactly the same, although the layout of the allocated store is the same. -- A. Kanawati [email]NO.antounk.SPAM (AT) comcast (DOT) net[/email] [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Luke Nowak Guest Posted: Fri Nov 19, 2004 4:53 pm    Post subject: Re: Question on memory allocation process of a 2D array KK wrote: Quote: Greetings, I am trying to see how the process of memory allocation is exactly acheived for a two dimensional array. Apparently, there are two cases I have to understand - static and dynamic. From the understanding I have so far, the process should be exactly the same in both cases. Here is the example that proves me wrong. /* Example to illustrate the processes of memory allocation of a two dimensional array */ #include 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, Well, in static case, the array elements are in the memory as follows .... array A [0][0] [0][1] [0][2] [1][0] [1][1] [1][2] in dynamic case: .... B array of char * [0] ... points to aaa [1] ... points to bbb .... then aaa [0] [1] [2] .... and then bbb [0] [1] [2] as you can see you allocate dynamicly an array of pointers, and then separate arrays of chars, so in memory they are put separately Greetings, Luke [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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