C Programming - Bitwise Operators - Discussion

I think whether in AND case both are true then only particle turned on. So I think OR case is the correct answer.

Anyone explain it.

I explain in simple manner

bit wise & truth table
=========================
a b z
=========================
0 0 0
1 0 0
1 1 1
0 1 0
==========================
only one condition satisfyied foe on bit excpt bit value is off position

Because && is a logical operator not a bitwise (bitwise is a operator that takes single bit at a time).

Which bit u want make 0 ...make only that bit 0 n allother bits 1...
ex:-to make 3rd bit 0 use &operator with one operand as11110111

~(1<<"bit position") is the operand

To make above operand you take a number 1 and then shift its only set bit to the number of the bit you want to turn off(1<<3) and then take AND with given number.

Bitwise 'AND' of any bit with zero bit is always zero (off).

Could anyone give an explanation in brief with example the above examples could not be understood

@Kumar

Let me explain.

How to turn off only the 4th bit (from right) in a 16-bit binary number?

unsigned int intFalg4Off = 0xfff7;
//Hex = 0xfff7 (or) Decimal = 65527 (or) Binay = 11111111 11110111

unsigned int intInputVal = 255;
//Decimal = 255 (or) 0x00ff (or) 00000000 11111111

unsigned int Result = intInputVal & intFalg4Off;

The result will be the & (AND operation) between the binary numbers given below :

11111111 11110111
00000000 11111111
00000000 11110111 (Decimal = 247 or Hex = 0x00f7)

Here 4th bit of the given input has been turned off. Therefore, intResult will contain the value 247.

Hope this will help you. Have a nice day!

Thanks sundar.