How does this output comes? I ask this question,because i found this question in C challenges...

Question

Please help me to find,How does this output comes.......
In these below codes.........

i.e int r = p - q; // output

You can also see that question in my post
https://www.sololearn.com/post/691377/?ref=app

https://code.sololearn.com/cxFx019iI7TW/?ref=app
https://code.sololearn.com/c9pBby9vD3oI/?ref=app

Answer 1

In the first code if you had done
r = a[0] - a[1]
The program would have printed
r = 1 - 2
I.e. -1

But you've done
r = p - q
So the program print the difference in sizeof(int) bytes (because they are arrays of int) between the position in memory of p and that of q.
Since array elements are by definition stored in contiguous memory location, 'locationOfSubscriptX' - 'locationOfSubscriptX+1' is by definition -1

In the second code the same thing is happening. You are subtracting the memory position of the pointers.
Since the result is 1, we know that, in the memory, a is stored right after b, I don't know why.

Answer 2

https://code.sololearn.com/cCEm3KJJivwy/?ref=app

Answer 3

https://code.sololearn.com/cLgixF4Ryq46/?ref=app

Answer 4

Yogeshwaran look at the hexadecimal addresses of the p and q in the second code.
I think we see different addresses, but the concept is the same.
For me they are exactly the same except for the last character: for a is 8 and for b is 4.
So when you do exaexaexa8 - exaexaexa4 you got 4. But since they are int and they are stored in 4 byte (in this case), 4 is a unit.
So you got 1.

If you change all the "int" types with "char" you'll see that the two addresses have a difference of 1 and not 4. Because char are stored in 1 byte.

Again I don't know why a is stored right after b. But that's another story.

Answer 5

Yogeshwaran
Have a look at this:
https://code.sololearn.com/cg0dY8X4JP5u/?ref=app

Answer 6

Yogeshwaran

Thank you Programmer Raja (தமிழன்) bro for yours effort in solving my doubt😊
can you tell me why these address size are changing......
while simply interchanging the variable initialisation.....
i.e &a > &b in one code and
&a < &b in another code
?

I think it depend on compiler

Answer 7

Thank you Davide 😊

Answer 8

Thank you so much Davide and Programmer Raja (தமிழன்) bro.....
For clearing my doubt.......

Now I only confused.which answer can be marked as a "best answer"😅

Because both answers are solved my doubt.....😊

Answer 9

Thank you Davide 😊.It make me to understand about pointer difference (subtraction in array).........
In Code 1.....

But I don't understand clearly how does output comes in code 2.......

Please answer related to code 2......
If any one know ,please answer.....

Answer 10

Bro it print one because the size of 1 integer is 4 bit
1 int =4bit
Let's have example

a=12(stored in address 30)
b=20(stored in address 34)

So the difference is 4bit which is equals to 1 integer
4bit =1 int

Answer 11

Thank you so much Programmer Raja (தமிழன்) bro😊
But &a - &b will give [ - 1 ] only bro from yours answer 😅

i.e
&a - &b
= address(30 - 34 ) = -4 bit[-1 int]
// am I right?

Then why Output comes as '1' 😅

It means that compiler made address of a is 34 && address of b is 30....
Or some other reason.....?

Please tell me bro......

Answer 12

Davide and Programmer Raja (தமிழன்)
See these code also......
Please explain it's behaviour...
https://code.sololearn.com/cv0pLwwdjl8K/?ref=app
https://code.sololearn.com/c9pBby9vD3oI/?ref=app

Answer 13

Yes Programmer Raja (தமிழன்) bro....😁😅.
But,
Please explain my comment bro...
By relating those two codes....

Answer 14

Thank you Programmer Raja (தமிழன்) bro for yours effort in solving my doubt😊
can you tell me why these address size are changing......
while simply interchanging the variable initialisation.....
i.e &a > &b in one code and
&a < &b in another code
?

Answer 15

Yogeshwaran you're welcome! 😄