Is This Statements Are Same And If Yes Then How?

Question

Is

Z = *(ptr1) * *(ptr2);

Equal to

Z = (*ptr1) * (*ptr2);

And

Z= *ptr1 * *ptr2 ;

Answer 1

As I understand it, the use of parentheses is only necessary when we use pointer arithmetic, as in if we do `*(ptr1 + 2) * *(ptr2 + 5)` because here we are dereferencing from another address rather than the one actually be pointed to by <ptr1> or <ptr2>.

(Edit)
The value of <Z> are the same for the 3 assignments above.

P.S. Specify language in Relevant Tags please.

Answer 2

When a pointer is used with array like that, & (address-of operator) is not necessary, because when an array is assigned to a pointer (<ptr1>), array <x> here is decayed into an address of the first element in the array <x> itself.

Read more about array decay in the response written by ~ swim ~ in the following thread:

https://www.sololearn.com/Discuss/2043862/?ref=app

Try this code (paste in main).

int x[5] = {1,2,3,4,5};

printf("Array <x> => %x\n", x);
printf("Address of array <x> => %x\n", &x);
printf("Address of first element in array <x> => %x\n", &x[0]);

int *ptr = x;
// array <x> decays into address of
// first element in it.
printf("Pointer <ptr> points to address => %x\n", ptr);

Answer 3

You're welcome Aman
BTW you can edit your reply or your original post above if you want. Sometimes autocorrect can be annoying I know : )

Answer 4

Thanks for adding language specification in tags 👍

Answer 5

Hii Ipang thankyou very much for explain me.. 😊 my doubles are clear now.

Answer 6

Sorry *doubts

Answer 7

Next Time I'll take care of this. 😋. Thank you.

Answer 8

Ipang understand

Int x[10];

int *ptr1 = x; is valid while x is array and size is 10

And int *ptr1 = &x; is invalid.

Whyyyy????

Answer 9

As I know & operator indicates address of variable!!!!