i=0 cout<< i++ << ++i << i; why the output is 122...shouldnt it be 022.?

Question

i think output should be 022...but why 1 is there?

Answer 1

It's strage isn't it. I think because they are in same cout. if you write it on multy cout like this:
int i=0;
cout<< i++ ;
cout<< ++i ;
cout<< i;
it will work fine.

Answer 2

What @Mohamed Aleid said is correct. Because i, i++, and ++i are all on the same statement, it's still going to run the pre-incrementations before anything else.

To prove this,

cout << i++ << ++i << ++i;

This would output 233, where as logically, you'd think the output would be 023.

It's just how the compiler does it.

Answer 3

122 is correct . because;
i++ one out = 0 but i=1
++i = 1+1 = 2 and i = 2
so; cout <<1 <<2 <<2;

Answer 4

the output will be 110 because in c++ the compiler calculates from right to left (in case of cascading of <<) and prints from left to right...:-)