C Programming - Structures, Unions, Enums - Discussion

In function data passing, expressions are evaluated from right to left,but printing results from left to right, so in the given problem first (i^j) evaluated,

i=0100
j=1000
i^j=1100 equivalent decimal is 12.

Then, next expression is (i|j&j|i)
the expression having bitwise AND and OR, in this bitwise AND having higher priority than bit wise OR, so grouping made as i|(j&j)|i, in this first (j&j) evaluated.

j=1000
j=1000
j&j=1000.

The remaining expression made as i|(1000)|i; here bitwise OR having two times that means same priority now we have to consider associativity. For bitwise operators left to right.

i=0100
1000
res=1100
1000

Final res=1100 decimal value 12.
third also same as the second expression so the result is 12.

Now printing the results in 12 12 12.

Basically ,meaning of AND operator is when 0&0 it returns 0 like this
0 0 =0
0 1=0
1 0=0
1 1=1

And for OR operator is
0 0=0
0 1=1
1 0=1
1 1=1

And for XOR operator is
0 0=1
0 1=0
1 0=0
1 1=0

Now our problem is i|j&j|i, i|j&j|i, i^j
So AND it has more precedence than all operator so it ll perform first,
i=4, j=8.

So its binary number is i=0100 n j=1000
So Step 1: 0100 | (1000 & 1000) | 0100

Becomes 0100 | 1000 | 0100
Step 2: (0100 | 1000) | 0100
Becomes 1100 | 0100
Step 3: (1100 | 0100)
Becomes 1100 which is equivalent to 12

Solution for i ^ j

Step 1: 0100 ^ 1000
Becomes 1100 which is equivalent to 12

Hence the answer 12,12,12.

Given data is int i=4, j=8;
i=4=0100
j=8=1000
|-> is bit wise OR
& -> is bit wise AND
^ -> is bit wise XOR
in sop(Structure Oriented Programing)Languages like c and oops(Object Oriented Programming)Languages like c++ and java the evaluation order is left to right

value of i|j&j|i

so first we have to calculate i|j
i=0100
j=1000
--------
1100=i|j
1000=j
---------
1000=i|j&j
0100=i
--------
1100=i|j&j|i=12
***************
next we have to calculate i^j
i=0100
j=1000
------
1100=i^j=12
so ans is 12,12,12

i= 4 =>0100 ; j=8 =>1000
solution for : i|j&j|i

& ->AND has highest Precedence than | ->OR operator hence should be evaluated first and then from left to right as usual .


So Step 1: 0100 | (1000 & 1000) | 0100
becomes 0100 | 1000 | 0100
Step 2: (0100 | 1000) | 0100
becomes 1100 | 0100
Step 3: (1100 | 0100)
becomes 1100 which is equivalent to 12


Solution for i ^ j

Step 1: 0100 ^ 1000
becomes 1100 which is equivalent to 12

Hence the answer 12,12,12

i=4,j=8;
step1:

i=4==>0010
j=8==>1000
=============
now we have to find i|j&j|i, i^j);
i|j&j|i==>0010 | 1000 & 1000 | 0100
solve & operator first
then we wil get
==>0100 | 1000 |0100
now solve from L -> R
==>1100 | 0100
==>1100
which is equal to 1*2^3+1*2^2+0*2^1+0*2^0
==>8+4+0+0
==>12
similarly remaing two also ................

4 value in binary is: 0100.

8 value in binary is: 1000.

4|8 = 0100|1000 = 1100 (According to the truth table of AND).

8|4 = 1000|0100 = 1100 (According to the truth table of OR).

12 value in binary is: 1100.

12 & 12 = 1100 & 1100 = 1100.

And 4^8 = 0100^1000 = 1100 = 12 (According to the truth table of XOR).

Do the bitwise operation from left to right

Binary form of 8 is 1000 and 4 is 0100

step:1 j=1000&1000=1000
step:2 i|j=0100|1000=1100
step:3 decimal value of 1100 is 12
step:4 do the same operation i|j&j|i value is=12
step:5 do the XOR operation i^j:0100^1000=1100=12
step:6 answer is 12 12 12

i=4 (binary-0100)
j=8 (binary-1000)
i|j&j|i...12(why)?
0100|1000=1100(by,bitwise OR operation)
1000|0100=1100(by,bitwise OR operation)
1100&1100=1100(12)(by,bitwise AND operation)

same as...i|j&j|i..12

and..
i^j...12(why)?
0100|1000=1100(by, bitwise XOR operation)
=12

Here &, | and ^ are bitwise operators,then operations will be performed on operands at bit level.

& - AND , | - OR , ^ - XOR operators

(4)d = (0000100)b, (8)d = (0001000)b

& has higher precedence over |

By considering all these, you will get the answer as expected.