### Discussion :: Bitwise Operators - Find Output of Program (Q.No.12)

Neetu said: (Aug 21, 2010) | |

4= 0100 8= 1000 it will give 4|8 = 1100 & 1100 = 1100 equals to 12 4^8 = 1100 = 12 |

Imran said: (Dec 2, 2010) | |

I am not understanding |

Vennila.P said: (Jan 11, 2011) | |

@Imran The solution is : Binary format of 4 = 0100 and 8 = 1000 Therefore 4|8 is = 0100|1000 = 1100 = 12 Similary 8|4 is also 1100 now i|j & j|i=1100 & 1100 =1100=12 Then i|j && j|i =12&&12 condition is true .. so it return 1. Then i^j = 0100^1000 = 1100 = 12. |

Roh said: (Jan 26, 2011) | |

Answer is correct but no clear explanation. So, can someone give detailed explanation? |

Madhureddy said: (Feb 5, 2011) | |

What vennila said is correct but you should perform & operation rather | operation. |

Santhu said: (Feb 28, 2011) | |

Vennila is correct. |

Wikiok said: (Mar 13, 2011) | |

& (bit AND) has in higher precedence than | (bit OR).So 4|8&8|4 == (4| (8&8) | 4) == ((4|8)|4) == (12|4) == 12 |

Atul Tailwal said: (Mar 25, 2011) | |

The solution is : OR(|) OPERATOR rules-> 1 1 =0, 1 0=1 AND(&) OPERATOR ->1 1 = 1, 1 0 = 0, 0 1= 0 Binary format of 4 = 0100 and 8 = 1000 Therefore 4|8 is = 0100|1000 = 1100 = 12 Similary 8|4 is also 1100 now i|j & j|i=1100 & 1100 =1100=12 Then i|j && j|i =12&&12 condition is true .. so it return 1. Then i^j = 0100^1000 = 1100 = 12. |

Raj said: (Apr 18, 2011) | |

This explanation is good |

Challenger said: (Aug 3, 2011) | |

@All. in i|j&j|i j&j wiill be evaluated first because of the order of precedence (& more than |) The answer will remain unchanged but that might not be the case in another situation. The right way is 0100|1000&1000|0100 (&) -> 0100|1000|0100 (leftmost|) 1100|0100 1100 12. |

Simhadri said: (Apr 24, 2012) | |

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 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 are becomes 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 (try it!) 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. Thats it. |

Saiprabha said: (Sep 7, 2012) | |

Here i=4 and j=8 now first i|j & j|i means 4|8 & 8|4 first 8 & 8 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 -------------------- 0 0 0 0 1 0 0 0 that means output is 8 again now 4|8 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 --------------------- 0 0 0 0 1 1 0 0 output is 12 now 12|4 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 ---------------------- 0 0 0 0 1 0 0 0 output is 12 the entire output for m is 12 there is only option which has 12 so the answer is C |

Neetu said: (Nov 23, 2013) | |

#include<stdio.h> #define x 50 #define y 60 int main() { int z = ++x + --y; printf("%d%d%d/n",x,y,z); } Can anyone explain me this program and its output? |

Rahul said: (Apr 20, 2014) | |

Bitwise & has more precedence comp then bitwise or why you first solve bitwise or please tell me. |

Mithi said: (Jun 12, 2014) | |

Does anybody know for sure whether: 1. & will be operated before | (as I too have read that precedence is (high to low) !, &, |) OR 2. | will be operated before & (because of left to right associativity). Kindly share your views. |

Sudha said: (Nov 6, 2014) | |

@Neetu. The output is: 51 59 110 The correct or not. Please anyone tell me. |

Sana said: (Aug 8, 2015) | |

51 59 110 is correct. |

Sana said: (Aug 8, 2015) | |

Can any one please tell me if i=4 and j=8 then how i&&j and i||j done? |

Aalapini said: (Sep 5, 2015) | |

Can some one explain i^j part? The function of ^ operator specifically. |

Tarun Ghosh said: (Sep 24, 2015) | |

^ this is binary XOR operator. Perform Bit-wise XOR operation Means odd no. of 1 is 1 otherwise 0. Then you will get your answer. |

Lokesh said: (Dec 21, 2015) | |

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. |

Anish Kumar said: (Aug 28, 2016) | |

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;. |

#### Post your comments here:

Name *:

Email : (optional)

» Your comments will be displayed only after manual approval.