When you assign the 16-bit value 0xffff
to the 32-bit integer b
, the variable b
actually becomes 0x0000ffff
. This means when you do the bitwise complement it becomes 0xffff0000
which is the same as decimal -65536
.
NOT operation on integer value
-
04-06-2022 - |
Question
I knew that ~ operator does NOT operation. But I could not make out the output of the following program (which is -65536). What exactly is happening?
#include <stdio.h>
int main(void) {
int b = 0xFFFF;
printf("%d",~b);
return 0;
}
Solution 2
OTHER TIPS
Assuming 32-bit integers
int b = 0xFFFF; => b = 0x0000FFFF
~b = 0xFFFF0000
The top bit is now set. Assuming 2s complement, this means we have a negative number. Inverting the other bits then adding one gives 0x00010000
or 65536
It is doing a bitwise complement, this output may help you understand what is going on better:
std::cout << std::hex << " b: " << std::setfill('0') << std::setw(8) << b
<< " ~b: " << (~b) << " -65536: " << -65536 << std::endl ;
the result that I receive is as follows:
b: 0000ffff ~b: ffff0000 -65536: ffff0000
So we are setting the lower 16 bits to 1
which gives us 0000ffff
and then we do a complement which will set the lower 16 bits to 0
and the upper 16 bits to 1
which gives us ffff0000
which is equal to -65536
in decimal.
In this case since we are working with bitwise operations, examining the data in hex gives us some insight into what is going on.
The ~ operator in C++ is the bitwise NOT operator. It is also called the bitwise complement. This is flipping the bits of your signed integer.
For instance, if you had
int b = 8;
// b in binary = 1000
// ~b = 0111
This will flip the bits that represent the initial integer value provided.
The result depends on how signed integers are represented on your platform. The most common representation is a 32-bit value using "2s complement" arithmetic to represent negative values. That is, a negative value -x
is represented by the same bit pattern as the unsigned value 2^32 - x
.
In this case, the original bit pattern has the lower 16 bits set:
0x0000ffff
The bitwise negation clears those bits and sets the upper 16 bits:
0xffff0000
Interpreting this as a negative number gives the value -65536
.
Usually, you'll want to use unsigned types when you're messing around with bitwise arithmetic, to avoid this kind of confusion.
Your comment:
If it is NOT of 'b' .. then output should be 0 but why -65536
Suggests that you are expecting the result of:
uint32_t x = 0xFFFF;
uint32_t y = ~x;
to be 0
.
That would be true for a logical not operation, such as:
uint32_t x = 0xFFFF;
uint32_t y = !x;
...but operator~
is not a logical NOT, but a bitwise not. There is a big difference.
A logical returns 0 for non-0 values (or false for true values), and 1 for 0 values.
But a bitwise not reverses each bit in a given value. So a binary NOT of 0xF
:
0x0F: 00000000 11111111
~0x0F: 11111111 00000000
Is not zero, but 0xF0.
For every binary number in the integer, a bitwise NOT operation turns all 1s into 0s, and all 0s are turned to 1s.
So hexadecimal 0xFFFF
is binary 1111 1111 1111 1111
(Each hexadecimal character is 4 bits, and F, being 15, is full 1s in all four bits)
You set a 32 bit integer to that, which means it's now:
0000 0000 0000 0000 1111 1111 1111 1111
You then NOT it, which means it's:
1111 1111 1111 1111 0000 0000 0000 0000
The topmost bit is the signing bit (whether it's positive or negative), so it gives a negative number.