Why output is 1 0 1, Please explain

What do you find difficult to understand?

Why 1 0 1 came as output

what do you think it should be

What does each line do?

what do you think it should be

arr1 arr2 arr3

Go line by line, what does each line do?

Go line by line, what does each line do?

1-3 Initialization of array, 4 declare a pointer where I store value of 3 array the using loop I am trying to print stored value of all three array

4 is wrong. That's not what it is

What does "int *parr[3]" mean?

What does "int *parr[3]" mean?

its a pointer

No

array of pointer

It's an array of pointer

Yes

yes

So, what is "{arr1, arr2, arr3}"?

idk

not understood

If parr3 is an array of pointers, what is "{arr1, arr2, arr3}"?

I'll rephrase the question. If you do printf("%d", arr1), what do you print?

With no brackets

stored value of in arr1?

No

What is arr1, without any bracket?

variable

What type?

int

No

why, its not int

Why output is 1 0 1, Please explain

its declared here int arr[]

Yes, it's declared with brackets, which means that, with brackets, it's an array of ints. If you take the bracket out, what type of variable you end up with?

int because it declared

With brackets, what happens when you take the brackets out in an array variable?

If you print the value of "arr1" and the value of "arr1[0]", what do you get in each case?

If you print the value of "arr1" and the value of "arr1[0]", what do you get in each case?

wait let me try

If you print the value of "arr1" and the value of "arr1[0]", what do you get in each case?

using arr1 it print adress

and second one print value

of 0th position

So arr1 is...

pointer?

Exactly

And when you initialize parr3, you are doing it with 3 variables of type...

not understand

beacuse when I print using arr1 it print some big number but with parr 3 it print 010

Step by step

If you print arr1, you'll get an address because it's a pointer

If you print parr[0], you'll get the same, but you are not printing parr[0]

What are you really printing?

If you print parr[0], you'll get the same, but you are not printing parr[0]

okay maybe i understood

wait

not understand

i am confused in why *parr[i] printing 1 0 1

Ok, we have arrived, that parr is an array of pointers

So parr[0] is a pointer

Which pointer?

Why output is 1 0 1, Please explain

here

Here what?

i am confused in why *parr[i] printing 1 0 1

.

What happens when you put * before a pointer?

What do you get with * before a pointer?

A duplet pointer

int a or int *b

or int* * duplet

A pointer pointing to a pointer

A duplet pointer

That's in a declaration. When you already declared a pointer, what do you get when you put a * before

A pointer pointing to a pointer

Not by far

You dereference it

Exactly

what?

pointer pointing to pointer?

No, forget about that

Not by far

?

pointer pointing to pointer?

If parr[i] is a pointer, what is *parr[i]?

pointer pointing to pointer?

.

What is the type of the array parr?

Please, don't confuse him

What is the type of the array parr?

Read the conversation from the beginning

yes so?

If parr[i] is a pointer, what is *parr[i]?

..?