c_programming

Question

What will be the output of the program ?

#include<stdio.h>

int main()
{
    int i=4, j=8;
    printf("%d, %d, %d\n", i|j&j|i, i|j&j|i, i^j);
    return 0;
}

    A.    12, 12, 12
    B.    112, 1, 12
    C.    32, 1, 12
    D.    -64, 1, 12

Comments

i got first 2
but my doubt is in 3rd
why we do bitwise xor
both are in integer
is it 4^6=4096 na

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no it is not power operator it is bitwise xor (^) operator   
for power we use mathematical power operator in math.h

Answer 1

Your Answer needs operator precedence table of C given http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm

so your code can be more specifically written as 

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

(j&j)=1000 AND 1000=1000=j(taking only lower 4 bits)

now the code become $i\mid j\mid i\Rightarrow \left ( i\mid j \right )\mid i\Rightarrow 1100\mid 0100= 1100$ (as "| " is left associative)

also $i\Lambda j\Rightarrow (1000 \Theta 0100)= 1100$(XOR)

so answer is 1100,1100,1100=(12,12,12)

A)

Comments

but why u consider ^ as bitwise xor not exponent?

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check the operator table ,the link i have given.

Who told that "^" operator=power in C,it is always bitwise XOR,

power in C is represented by pow(base,power)

"^" operator=power is in ENGLISH not in C

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@Sourav Sir , but in "now the code become i∣j∣i⇒(i∣j)∣i⇒1100∣1000=1100 (as "| " is left associative)"

here in last bitwise OR operation, you wrote the value of 'j' instead of 'i....here i=4=0100  ... please correct me if I'm wrong

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Question-:i|(j&j)|i

step 1:j&j=j

step2:i | j | i

step 3:( i | j )| i

step 4:1100|1000=1100.

Now say where is your doubt?

---

step 3:( i | j )| i

step 4:1100|0100=1100.

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updated ..!!! thank you @manoj and @ankit.@ankit don't call me sir ,as i am also aspirant like you :)

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ok :)

Answer 2

ans will be A  12,12,12,

| bitwise or

& bitwise and

^ bitwise xor

now i=4 in bits 100  , j=8 in bits 1000   j&j =1000 or 8 so i | j&j= 4|8=12|4=12

so first 2 values will be 12 and 12

i^j=0100 xor 1000 =1100 so it is also 12

Comments

but why u consider ^ as bitwise xor not exponent?