CODEWITHSUNDEEP
c
Shaw Ankush
Shaw Ankush

Reputation: 113

Initialization of pointer to array

I have learnt that the name of the array is actually the address of array_name[0]. Then why does is require to add ampersand sign before the name of the array while initializing a pointer to the array. 

  int (*pointer_name)[5] = &array_name;

I have tried:

  int *pointer_name = array_name;

and it works fine. What is the difference between the two other than the "type of pointer"? And also what are the pros-cons of either of them. When to use them? Does anyone of them has got any greater/ better functionality over other?

Upvotes: 3

Views: 73

Answers (3)

Vlad from Moscow
Vlad from Moscow

Reputation: 310930

The type of the object pointed to by the pointer declared like

int (*pointer_name)[5] = &array_name;

is int[5] . This means for example that this operator

sizeof( *pointer_name )

yields the value equal to 5 * sizeof( int ). And that if to use the pointer arithmetic as for example pointer_name + 1 then the address obtained by this expression will be equal to the address stored in the pointer plus value of 5 * sizeof( int ).

The type of the object pointed to by the pointer declared like

int *pointer_name = array_name;

is int . This means for example that this operator

sizeof( *pointer_name )

yields the value equal to sizeof( int ). And that if to use the pointer arithmetic as for example pointer_name + 1 then the address obtained by this expression will be equal to the address stored in the pointer plus value of sizeof( int ).

A pointer declared like this

int (*pointer_name)[5];

usually used for two-dimensional arrays that points to "rows" of array.

For example

int array_name[2][5];

int (*pointer_name)[5] = array_name;

// outputs 20 provided that sizeof( int ) is equal to 4
printf( "%zu\n", sizeof( *pointer_name ) ); 

// outputs 4 provided that sizeof( int ) is equal to 4
printf( "%zu\n", sizeof( **pointer_name ) );

pointer_name points to the first row of the array array_name. pointer_name + 1 points to the second row of the array.

Upvotes: 0

Jean-Baptiste Yunès
Jean-Baptiste Yunès

Reputation: 36391

int *pointer_name = array_name;

declares a pointer to int that points to the first int of the array array_name.

int (*pointer_name)[5] = &array_name;

declares a pointer to an array of 5 ints that points to the array array_name.

Addresses are the same but types not.

If you use pointer arithmetic on those you have, in the first case:

pointer_name++;

will just point to the second int of the array, while in the second case it will point just after the whole array.

Upvotes: 3

P.P
P.P

Reputation: 121357

Then why does is require to add ampersand sign [..]. I have tried : int *pointer_name = array_name; And it works fine.

Because the types are different.

  • &array_name is a pointer to an array of 5 ints and has type: int (*)[5].

  • array_name gets converted into a pointer to its first element when you assign it to pointer_name (which is equivalent to &array_name[0]) and has type: int*.

If array_name is an array of 5 ints then both:

int (*pointer_name)[5] = &array_name;

and

int *pointer_name = array_name;

are valid. Just how you'd access them later through these two pointers is different.

Upvotes: 2

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