C Programming - Bitwise Operators - Discussion

~ is a unary operator.

Then it should have operated on a and saved the result in the same variable. But why isn't it happening?

Can anyone explain?

int i=4, j=8;

printf("%d %d %d\n",i|j&j|i,i|j&&j|i,i^j);

Can some explain me the way above question going to work.

@Deepak.

1st (i|j&j|i).
Where i = 4= 0100(binary) and j = 8 = 1000(binary).
Now i|j where'|' is OR operator i.e it will 1100(12).
Similarly j|i will be 1100(12).

Now i|j&j|i where '&' operator as AND operator
1100(i|j)
& 1100(j|i)
1100(i|j&j|i) i.e 12.

Now second(i|j&&j|i)

Hence (12&&12) where '&&' is logical operator which output is either 0 or 1.

Hence here Rvalue = Lvalue so it is true.
is 1. so print 1.

Now 3rd (i^j) where '^' is XOR operator
i = 4 (0100)
^ j = 8 (1000)
(i^j) 1100.

So output is as 12 1 12 respectively.

Here ~a is used but not assign a value back to a so that a will not be affected.

What is the meaning of 0x..?

What does 0Xfff represents? can anyone explain clearly?

Like other unary operator they store the resultant value in operand itself on which the unary operator is applied.

For example a =7 and ++a stores 8 in a so why doesn't ~a stores the result in a.

Thank you @Taruna.

I satisfy with you answer. What about two byte compiler?

a = 1111 1111 1111 1111

Let b =~ a, means b = 0000 0000 0000 0000

[
Concept : take composite resultant value is b = (-1)*(a+1).
Here 'a' may be +/- in the form of Decimal value.
]

From above eg:

a = 0xffff.

a = -1 [decimal value of 0xffff].

So due to our concept b =~a => b = (-1)*(-1+1) therefor resultant of b = 0

b = 0 [means hexadecimal value is 0000 0000 0000 0000].

But here from above example:

Line 1: a = 1111 1111 1111 1111

Line 2: ~a = 0000 0000 0000 0000

But value of 'a' does not change in line 2.

So a = 0xffff.

Here a = 0*ffff.
Which can also be written as 0000ffff.

So ~a = ffff0000.
Therefore output will be ffff.