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carrayspointersimplicit-conversion
shubh
shubh

Reputation: 1

what does int (**p)[2] declare ? And what would be stored in array **p?

And what would be stored in array **p?

int main()
{
    int i, a[2][2]={{1,2},{3,4}};
    int (**p)[2];
    p=a;
    for(i=0;i<4;i++)
        printf("%d  ", *(*p+i));

    return 0;
}

Upvotes: 0

Views: 229

Answers (3)

Nigel Winterbottom
Nigel Winterbottom

Reputation: 46

The questioner's code does not compile:

int i, a[2][2]={{1,2},{3,4}};
int (**p)[2];
p=a;

error C2440: '=': cannot convert from 'int [2][2]' to 'int (**)[2]'

If the intention was to define a pointer to the 2x2 array a, then p should be defined as a pointer to array 2 of array 2 of int and used thus:

int main()
{
    int a[2][2] = { {1,2},{3,4} };
    int(*p)[2][2];

    p = &a;
    printf("{%d, %d} \n", (*p)[0][0], (*p)[0][1]);
    printf("{%d, %d} \n", (*p)[1][0], (*p)[1][1]);
    return 0;
}

The output is:
{1, 2}
{3, 4}

Upvotes: 0

chux
chux

Reputation: 153447

what does int (**p)[2] declare ?

p as pointer to pointer to array 2 of int
(Very useful site C gibberish ↔ English)

what would be stored in array **p?

In **p an array 2 of int.

p is a pointer. p would also store a pointer. That stored pointer in p points to an array 2 of int.

  int array2[2] = {5, 7};
  int (*pointer)[2] = &array2;
  int (**p)[2] = &pointer;
  printf("%d %d\n", (**p)[0], (**p)[1]);  // output "5 7\n"

Upvotes: 3

Vlad from Moscow
Vlad from Moscow

Reputation: 310960

I think there is a typo in this declaration.

int (**p)[2];

There must be

int (*p)[2];

That is it is a declaration of a pointer to array of the type int[2]

And in this assignment

p = a;

the array a is implicitly converted to pointer to its first element that has the type int[2]. That is in this assignment the right side hand operand has the type int ( * )[2] and the left side hand operand has the same type.

Otherwise you will get a compilation array because the pointers int ( ** )[2] and int ( * )[2] are not compatible,

In this for loop

for(i=0;i<4;i++)
    printf("%d  ", *(*p+i));

there is in fact reinterpretation of the pointer to the type int *. As the pointer p has the type int ( * )[2] then dereferencing it you get lvalue of the type int[2] but used in the expression *p+i this array in turn is converted to pointer to its first element. That is inside the expression *p+i you have pointer to the element a[0][0].

You could rewrite the expression

*(*p+i)

like

( *p )[i]

If to use a pointer of the type int( ** )[2] then the body of the function main could look like

int i, a[2][2]={{1,2},{3,4}};
int (*q)[2] = a;
int ( **p )[2] = &q;

for(i=0;i<4;i++)
    printf("%d  ", *( **p +i ) );

That is one more indirection of accessing elements of the array is added.

Upvotes: 0

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