In my class i found assigning an array of 4 integer elements to a pointer in the below fashion...i did not get how is this possible
it should be some thing likeCode:int a[4]; int (*p)[4]=&a;
"&a" is nothing but "a" itself right which stores the base address of the array. Then whats the meaning "int (*p)[4] = &a".....can it be "int (*p)[4]=a" will bothe give the same resultCode:int *p=a;
&a is the address of a, which is a pointer. Therefore &a is a pointer to a pointer, which is what int (*p)[4] declares.
Last edited by Dave_Sinkula; 10-24-2006 at 09:42 PM. Reason: ...
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
That means u say that the below declaration is perfectly right ?
but whats the significance of 4 there...if we give 7 there will this program work..?Code:int a[4]; int (*p)[4]=&a;
I imagine you tried that and your compiler at least gave you a warning.Originally Posted by enggabhinandan
A pointer to an array of 4 ints should not be happy with initialization to an array of 7 ints.
Arrays have a size. So a pointer to an array must know the size.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
