help with confusing declarations please

[csepraveenkumar]

in k&r c the following declaration is given to be pointer to array[13] of int

int (*daytab)[13]

on similar lines i initialize a pointer to array[2]of int in my program as follows

int c[2];
int (*d)[2];
d=c;

when compiling this code i get a warning as follows

sizeof.c: In function ‘main’:
sizeof.c:7: warning: assignment from incompatible pointer type

what is generating this warning and how to get rid of it? also are there any good resources for understanding complicated declarations in c?

[CommonTater]

Actually you could just use

int c[20];
int *d;
d = c;

There's no need to make it any more complex than it has to be.

Also note that depending which edition of K&R you have and the compiler you're working with, some things may not be entirely compatible between the two.

[csepraveenkumar]

thank you for your reply. what you sad can be done. but i am trying to understand what that declaration means.

another important observation. when i do the following initialisation the the warning disappears

int c[2];
int (*h)[2];
h=&c;

please help in understanding this declaration.

[Bayint Naung]

int (*daytab)[13];
int days[2][13] ={  {0 , 1,2,3,4,5,6,7,8,9,10,11,12 },
                            {12,11,10,9,8,7,6,5,4,3,2,1,0 } };
daytab = days;   // points to first element &days[0] { 0,1,2,....}
daytab + 1        // points to next element( array {12,11,10...})

daytab is pointer to array[of 13 int].

[Adak]

thank you for your reply. what you sad can be done. but i am trying to understand what that declaration means.

another important observation. when i do the following initialisation the the warning disappears

int c[2];
int (*h)[2];
h=&c;

please help in understanding this declaration.

int c[2] has two elements, and is an array of int's. (it helps to read them from right to left, instead of left to right).

int (*d)[2] has two elements, but is an array of two pointers to int. So they can't be set equal to each other - they're different beasties. Edit: See the correction below, this one I got wrong.

But you can set the pointer to the address of the base of the int array.

[Salem]

> int (*d)[2] has two elements, but is an array of two pointers to int.
Pointer to an array of 2 ints.

int *d[2] is an array of 2 pointers to int.

The () are important here.

[Adak]

Thank you, Salem. The (*d)[2] declaration is one I've never used.

[Salem]

You've never passed a 2D array to a function before?

As a function parameter, you can write

void foo ( int arr[2][2] );
void foo ( int arr[][2] );
void foo ( int (*arr)[2] );

But you can only use the last form to declare a variable with the same type.

[Abhinav Arora]

thank you for your reply. what you sad can be done. but i am trying to understand what that declaration means.

another important observation. when i do the following initialisation the the warning disappears

int c[2];
int (*h)[2];
h=&c;

please help in understanding this declaration.

Hello Praveen

 int (*daytab)[4];

This is a pointer to an array of 4 integers.

Now if you have the following :

int c[2];
int (*d)[2];
d=c;

Now, the name of an array always gives us the address of the first element of the array.
So the 'c' here is the Integer Pointer Ito the first int in c. On the other hand 'd' is a Pointer to an Array of 2 Integers. It is because of this reason that you get a Warning.

On Incrementing an Integer pointer, you get a pointer to the next int, however on incrementing the pointer to array, you jump over the entire array and point to the next array of the same dimension.

Now coming to this code:

int c[2];
int (*h)[2];
h=&c;

As we all know that the name of the array represents the pointer to the first element of the Array. However this rule has some exceptions:

1. When you use sizeof(arr) , then it does not return the size of the pointer, instead it returns the size of the array. In your example, the answer will be 8 (because c is an array of 2 integers) and not 4(which is the size of an integer pointer)

2. Similarly when the name of the array is used with the '&' Operator, it converts the pointer to the entire array.

Hence when you use

h=&c

Here, &c is now the pointer to the array of 2 ints and is of the same type as h and thus you do not get the warning.

The code can also be written as

h= (int (*)[2])c;

Here we have typecasted c into a pointer to an array of 2 ints.

Hope this was helpful.

[laserlight]

Abhinav Arora, that is a good explanation. However, the discussion in this thread has since died out, and there was no errors that your contribution corrected.

*thread closed*