C Programming - Bitwise Operators - Discussion

For i|j&&j|i or operator has high precedence so j|i=12 and i|j=12.

Result of i|j && result of j|i is true because in && operator it checks both operands are non zero.

So it produces 1 as its output.

An unary operator has higher precedence than arithmetic operator so why pre-increment and pre-decrement will happen first ++x + --y i.e. 51 + 59 = 110.

x=50 ++x=51;.

| what is this symbol means?

@Sana.

X and y values are defined as constant.
So values don't change.
The answer is 50 60 110.
If it is wrong pls tell me how it comes?

4 = 0100
8 = 1000

Let taken 1=true and 0=false.

| means "or" operator i.e. only false to false condition become false but in other cases become a true condition.
& means "and" operator i.e. only true and true condition become true but in other cases it becomes false.

^ means "double implication" i.e. true, false condition and false, true conditions become true but in other cases it shows false.

Then,
i|j = 0100 | 1000 = 1100 = 12
j|i = 1000 | 0100 = 1100 = 12
and then i|j&j|i = 1100 & 1100 = 1100 = 12
So first condition prints 12
And third condition also prints 12.

In the second condition following '&&' is operator shows true prints 1 or false prints 0 for the following condition. So 12&12 = 12 so it is true so it prints 1.
That's it.