Some one explain this code plz,^^

Question

pointers and "&"
https://code.sololearn.com/cg72o4hzX0gC/?ref=app

Answer 1

yes you are right KrOW as I set a code if you try it in turbo cpp4 u will get answer as 1 0 1 1 1 but in gcc and in our play ground will get answer as 1 1 1 1 1

Answer 2

problems like this code the main thing is priority and associativity matters as Deepak sir said I have one more code which suits the same problem
https://code.sololearn.com/cu7gKp7uNqOX/?ref=app

Answer 3

++i is pre increment and it gets executed first.. which makes i as 2.

I** is post increment and it has second priority after pre increment.. i and *p is same... so (*p)++ is 2 only and then value updates to 3. so output is 322

Answer 4

Anas Sanba int *p is a pointer variable.
Its assigning the value of i variables memory address -
int *p = &i;

Revise the tutorial.
Or, go to
https://youtube.com/user/thenewboston

Search for pointers.

Answer 5

Guys maybe i am not been clear.... Order evalution of expression passed like function parameters ARE NOT DEFINED by standards then ANY compiler can evalutate they in ANY order.... Here for more simple words https://www.geeksforgeeks.org/g-fact-20/

and here from official source https://en.cppreference.com/w/c/language/eval_order

Answer 6

I want make note that evalutation order with side-effect expressions its not specified in standards (or they have undefined bahviour eg. implementation dependents)

https://www.geeksforgeeks.org/g-fact-20/

In practice dont assume that an parameter its evalutated before another

Answer 7

Anas Sanba what shady business is going on! Why is the output messed up. I don't get it.

Anyway:
Pointers points to the memory address of a variable.
The & operator is used to indicate the memory.

Answer 8

Deepak Gautam shouldn't that be 1? for the first one.

Answer 9

KrOW ,Thank you. I was really confused bout that.

Answer 10

thank you so much guys i get it now 😙

Answer 11

as p is pointing to i, *p is nothing but value pointed by i

Answer 12

Deepak Gautam

Answer 13

it's juste one?

Answer 14

ah!! there is exchange between p and i ?

Answer 15

explain me int *p =&i;?

Answer 16

how *p=3?

Answer 17

Ketan so *p =i=1?

Answer 18

I did know about precedence but this worked very weirdly.

I am but confused.

Answer 19

As krow mentioned it's behavior isn't specified in the documentation. I thought arithmetic operations in printf didn't have effect on the variables but in this case it clearly does.

Answer 20

deepak why value of i become 3?