C Programming - Bitwise Operators - Discussion

Calculate: -1^~0

-1 => 1111 1111 1111 1111
~0 => 1111 1111 1111 1111
--------------------------------
-1^~0 => 0000 0000 0000 0000 => is binary of 0 (zero).
--------------------------------

Firstly find the binary of -1:
1 => 0000 0000 0000 0001
1's compliment of 0000 0000 0000 0001 is:
All 0 is converted into 1 and all 1 into 0, now

1111 1111 1111 1110

Now 2's complement of 1111 1111 1111 1110 is:
Add 1 into 1's complement
1111 1111 1111 1110
+ 0000 0000 0000 0001
----------------------------
1111 1111 1111 1111 => is binary of -1
----------------------------
Now, binary of 0 is:
0 => 0000 0000 0000 0000 => is binary of 0
~0 => to convert all 0 into 1 and all 1 into 0, i.e.
1111 1111 1111 1111 => is binary of ~0
-----------------------------------
^ is called Bitwise XOR operator
0 ^ 0 = 0
0 ^ 1 = 1
1 ^ 0 = 1
1 ^ 1 = 0
------------------------------------

-1 means it will treated as 2's complement
that means adding 1 to the result of 1's complement.
Binary form of 1 is :
0000 0000 0000 0001
1's :---- 1111 1111 1111 1110
adding 1 then
:----- 1111 1111 1111 1111



and ^ means it will print 0 if both are same(1,1---0)(0,0----0)
otherwise 1 (1,0 -------1)
0's is :-- 1111 1111 1111 1111


now for -1 **** 1111 1111 1111 1111
for ~0 **** 1111 1111 1111 1111
now we will perform the ^ operaiton
1111 1111 1111 1111
1111 1111 1111 1111
-------------------------------------
0000 0000 0000 0000
so the answer is zero

Let you take int i=1;

int is 2-byte range in turbo c, 4 bytes in gcc, now consider turbo c==>2 bytes means 16 bits right so these 16 bits stored in memory for 1 is (0000 0000 0000 0001). Before discoursing about -1; what is 2's complement of x, that is exactly equal to -x. note: negative numbers stored in the memory is 2's comp of a positive number. so we have to find what is 2's comp of 1. 2's comp=1'comp+1;

0000 0000 0000 0001==>1'comp is==>1111 1111 1111 1110 ==>add 1==>1111 1111 1111 1111. So finally the -1 is stored in memory is 1111 1111 1111 1111.

After the First macro function executed

#define P printf("%d\n", -1^~0);
#define M(P) int main()\
{\
P\
return 0;\
}
M(printf("%d\n", -1^~0);)

After the second Macro function executed

#define P printf("%d\n", -1^~0);
#define M(P) int main()\
{\
P\
return 0;\
}

int main()
{
printf("%d\n", -1^~0);
return 0;
}

If an integer on this machine is 2-bytes, why is the answer not 32,766?
0000 0000 0000 0001=1
1111 1111 1111 1110=-1 1's compliment of -1

1111 1111 1111 1111=-1 in 2's compliment
^1111 1111 1111 1111=0 after negation or inverting
........................
0000 0000 0000 0000
so therefore we know that 0^0=0
1^1=0
so the answer is 0(Zero).
thanks frds.
all the best.

00000000 00000000 00000000 00000001 = 1
11111111 11111111 11111111 11111110 = -1 in 1's compliment
we now have to add 1 to make it into 2's compliment .
So,

11111111 11111111 11111111 11111111 = -1 in 2's compliment

11111111 11111111 11111111 11111111 (0 after negation or inverting)

for 0^0 = 0
1^1 = 0

So the answer is 0 (zero).

main() is function first it will call , after that printf will work.

Note : Printf function you can write in inside main function.

Like main(printf("hai"))

If anybody ask Printf function how can write without ends with semicolon means this is solution.

Most of the part is correctly explained.
Except ~ is 1s complement not negation

Negation of 0 is negation is logical operation
1s complement is bitwise operation

Negation of 0 will yield 1

But
~0= 1111 1111 1111 1111

Answer = 0 because,

Binary of -1(in 2 byte) 1111 1111 1111 1111.
Binary of ~0(in 2 byte) 1111 1111 1111 1111.
Now, apply Xor (-1 ^ ~0) = 0000 0000 0000 0000(ans).

Hint:use Xor truth table.

\ split the macros into multiple line
For example

#define P(int i) { \
printf("%d",i);\
}

Or

#define P(int i) { printf("%d",i);}