Published by
Jan 14, 2008

Array is not pointer

Score: 3.1/5 (23 votes)
*****
With many thanks for these useful tutorials, I felt it's necessary to send this text about pointers and arrays. Unfortunately pulling out something wrong that is put in humans head is a bit difficult. So understanding the things correct and precise is very important to avoid further misconceptions.

An array is not equal to a pointer. It is a sequence of simple variables in memory.

When we write 
1
2
 
int array[3];
array[2]=666;


C/C++ compiler doesn't see array[0] as an address to an integer value, it takes it directly as a value, exactly as same as writing
1
2
 
int var;
var=66;


That's obvious that "var" is not a pointer exactly as array[2] is not.

But if we use a pointer instead of an array, the face of the code is the same but compiler generates different assembly code. For example 
1
2
 
int *ptr = new int[3];
ptr[2] = 66;


Is similar to the first code but not with the same meaning to the compiler. In the first code (second line), compiler generates code that will do the following:

1) Go two places next of the array[0] and make it equal to 666.

But in code with pointer it is:
1) Fetch the value (address) of the ptr[0].
2) Add two to it.
3) Make the value pointed by it to 66.

Actually the value of "array", "&array" and "&array[0]" is equal. But the type of "&array" is different (a memory address to an array not an array member).

Here is another example to make the article more understanding. I want to write a program that gets an integer from user, adds 4 to it and then prints out the result. Once I write it using an integer pointer and once with an integer variable.
With integer it will be:
1
2
3
4
5
6
7
 
#include<iostream>
main(){
    int int_input;
    cin>>int_input;
    cout<<(int_input + 4)<<endl;
    return 0;
}


And using a pointer it will be:
1
2
3
4
5
6
7
8
 
#include<iostream>
main(){
    int *int_ptr = new int[1];
    cin>>*int_ptr;
    cout<< (*int_ptr + 4)<<endl;
    delete(int_ptr);
    return 0;
}


Who thinks these programs are exactly the same?
Lets take a look at their assembly. For the first code with an integer it is:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
 
2212: main(){
00401000   push        ebp
00401001   mov         ebp,esp
00401003   sub         esp,44h
00401006   push        ebx
00401007   push        esi
00401008   push        edi
2213:     int int_input;
2214:     cin>>int_input;
00401009   lea         eax,[ebp-4]
0040100C   push        eax
0040100D   mov         ecx,offset cin (00414c58)
00401012   call        istream::operator>> (0040b7c0)
2215:     cout<<(int_input+4)<<endl;
00401017   push        offset endl (00401070)
0040101C   mov         ecx,dword ptr [ebp-4]
0040101F   add         ecx,4
00401022   push        ecx
00401023   mov         ecx,offset cout (00414c18)
00401028   call        ostream::operator<< (0040b3e0)
0040102D   mov         ecx,eax
0040102F   call        ostream::operator<< (00401040)
2216:     return 0;
00401034   xor         eax,eax
2217: }


And for the code with pointer it is:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
 
2212: main(){
00401000   push        ebp
00401001   mov         ebp,esp
00401003   sub         esp,4Ch
00401006   push        ebx
00401007   push        esi
00401008   push        edi
2213:     int *int_ptr = new int[1];
00401009   push        4
0040100B   call        operator new (004011b0)
00401010   add         esp,4
00401013   mov         dword ptr [ebp-8],eax
00401016   mov         eax,dword ptr [ebp-8]
00401019   mov         dword ptr [ebp-4],eax
2214:     cin>>*int_ptr;
0040101C   mov         ecx,dword ptr [ebp-4]
0040101F   push        ecx
00401020   mov         ecx,offset cin (00414c38)
00401025   call        istream::operator>> (0040b8a0)
2215:     cout<< (*int_ptr + 4)<<endl;
0040102A   push        offset endl (004010a0)
0040102F   mov         edx,dword ptr [ebp-4]
00401032   mov         eax,dword ptr [edx]
00401034   add         eax,4
00401037   push        eax
00401038   mov         ecx,offset cout (00414bf8)
0040103D   call        ostream::operator<< (0040b4c0)
00401042   mov         ecx,eax
00401044   call        ostream::operator<< (00401070)
2216:     delete(int_ptr);
00401049   mov         ecx,dword ptr [ebp-4]
0040104C   mov         dword ptr [ebp-0Ch],ecx
0040104F   mov         edx,dword ptr [ebp-0Ch]
00401052   push        edx
00401053   call        operator delete (00401120)
00401058   add         esp,4
2217:     return 0;
0040105B   xor         eax,eax
2218: }


19 lines Vs 32 lines. Therefore, you see, an integer is different to a "pointer to integer". An integer is a place of memory where an integer number is kept but an integer pointer (pointer to integer) is a place of memory where an address is saved. Compiler knows that is an address of a place of memory where an integer is held. I do not explain the assembly code since this article is for beginners and I want to keep it short.

As I said array is a sequence of variables in memory. In the example above, I concluded that a pointer is not an integer variable, so it is clear that it cannot be a sequence of them too.

Please feel free to send comments and your idea about the article.