c/c++ pointer to an array vs pointer to a pointer

Question

I thought array and pointer are basically the same thing, until I run this program:

int main() {
  int* a = new int(19);
  int b[1];
  b[0] = 19;

  printf("*a: %d\n a: %p\n &a:%p\n", *a, a, &a);
  printf("*b: %d\n b: %p\n &b:%p\n", *b, b, &b);
  delete a;
}

output is:

*a: 19
 a: 0x7f94524000e0
 &a:0x7fff51b71ab8
*b: 19
 b: 0x7fff51b71ab4
 &b:0x7fff51b71ab4

can someone please explain why the output of &b is the same as b?

Thanks! -Erben

This kind of arrays refers to a dead-memory in your program, and is different from pointers. Pointers can refer to any adress. Another thing: if you use int tab[n][m], you can't use it with a int **ptr but with a int *ptr -> compiler will replace calls to tab[i][j] by tab[i * m + j]. — anotherdev
– anotherdev, Commented Nov 2, 2013 at 19:56

Dietmar Kühl · Accepted Answer · 2013-11-02 19:58:14Z

6

Well, b is an array. Under the slightest excuse it will decay into a pointer to the first element of b. Note, however, that the expression b and &b are not equivalent: b decays into a pointer the first element, i.e., it is of type int* while &b is a pointer to the array, i.e., it is of type int(*)[1].

answered Nov 2, 2013 at 19:58

Dietmar Kühl

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3 Comments

Silent Control Over a year ago

But why are a and &a different?

Dietmar Kühl Over a year ago

@Robertfrost: a and &a are clearly to different entities: a is an object which happens to be a pointer to an int and &a is that object's address, i.e., it is a pointer to a pointer to int. Arrays are special, though: if you just name and evaluate them, they automatically turn into a pointer to their first element, i.e., b is equivalent to &b[0]. This element lives at the same address as the start of the array object which is obtained from &b.

John Bode Over a year ago

@Robertfrost: See this answer for some historical context.

masoud · Accepted Answer · 2013-11-02 20:18:10Z

Arrays and pointers are not the same. A pointer can behave like an array (e.g. accessing by index).

&b is a pointer to the whole array and b is a pointer to the first element. They may point to a same address in memory but they are totally different things.

        +-------------------------------+  
        |+-----+-----+-----+-----+-----+|  
        ||     |     |     |     |     ||  
 &b---->||  0  |  1  |  2  | ... |  N  ||  
        ||     |     |     |     |     ||  
        |+-----+-----+-----+-----+-----+|  
        +---^---------------------------+  
            |                              
            b

a and &a are two different values because one is the address of that "0", one is the address of that "a". but in b, you only have one box and therefore one memory address.

Jaywalker · Accepted Answer · 2013-11-02 20:20:44Z

-1

a is a variable. You are allocating memory using new and assigning the result to this pointer. You might decide to store something else in a later in your program.

b is different. It's not a variable in the sense that it cannot store different addresses. It's an array, having a fixed start location.

Thus, b and &b are same. But the contents of a and the actual address of a are different.

answered Nov 2, 2013 at 20:20

Jaywalker

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