正数和负数的按位运算符

Question

    -5 / 2 = -2

    -5 >> 1 = -3

I learn from my teacher that >>1 divides the number by 2. It works on positive number but it does not work on negative numbers. Can someone explain to me??

Thanks

Answer 1

As BЈовић & mystical states, using bit shift operators on negative numbers is implementation defined.
The reason for this is C doesn't distinguish between logical and arithmetic bit shifting.
(Arithmetic pads with the most significant bit, logical pads with 0's)
for positive numbers this doesn't matter, for both arithmetic and logical bit shifts would keep the most significant bit as a 0:
Arithmetic 5>>1
0000 0000 0000 0101 = 5
to
0000 0000 0000 0010 = 2

Logical 5>>1
0000 0000 0000 0101 = 5
to
0000 0000 0000 0010 = 2

however with a negative number (2's comp)
Arithmetic -5>>1
1111 1111 1111 1011 = -5
to
1111 1111 1111 1101 = -3

Logical -5>>1
1111 1111 1111 1011 = -5
to
0111 1111 1111 1101 = 32,765

or at least, this is how i understand it

Answer 2

It works on positive number but it does not work on negative numbers.

Using shift operator on negative integer numbers is implementation defined.

---

[expr.shift]/3 tells this :

The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a non-negative value, the value of the result is the integral part of the quotient of E1/2E2 . If E1 has a signed type and a negative value, the resulting value is implementation-defined.

Answer 3

I learn from my teacher that >>1 divides the number by 2.

It doesn't divide the integer by two, but it performs (depending on the value) a logical or an arithmetic shift by one bit to the right. It happens to be equal to a division by two under some circumstances.

It works on positive number but it does not work on negative numbers.

It works in both cases, but the exact behavior is not mandated by the standard, but rather implementation-defined. It usually divides by two and truncates the result towards negative infinity, in constrast to towards zero as a normal division would do.

For reference:

• http://en.cppreference.com/w/cpp/language/operator_arithmetic

Answer 4

First of all,
5 in binary is 0000 0000 0000 0101 but what about -5 ? Here it is :

1. Change 1 to 0 and 0 to 1 ,then we get 1111 1111 1111 1010
2. Then take that number + 1 , we get 1111 1111 1111 1011

Now we get: -5= 1111 1111 1111 1011 ( it's in 2's complement form)
So here is how to calculate -5>>1 :

1. Move each bit of -5 from left to right ( >> ) we get 111 1111 1111 1101 (only 15 bits left)
2. Because -5 is negative number, so we have to fill '1' to the first bit to make it become 16 bits Then we get 1111 1111 1111 1101 ( it's still in 2's complement form)
3. Now convert it to a normal binary form by change 0 to 1 , 1 to 0 (except the first bit because it defines negative number in 2's complement), then plus '1' . so we get 1000 0000 0000 0011 = -3

Answer 5

I think answer is correct. As '/' (division) operator generate quotient (result of division).

In your problem :

-5/2 = -3(quotient) and 1(remainder ). 

So this is ok with both positive and negative number.

positive Number:

5/2 = 2(quotient) and 1(remainder ). 

So it is fine with positive Number.

NOTE

Remainder never be a negative number. It is always positive Number.

Answer 6

I guess the answer to -5>>1 = -3. In case of a positive number, say 5, division by 2 gives 2.5 rounding off to the nearest smallest integer ie 2

But when we consider a negative number, -5, division by 2 gives -2.5. Its rounding off to the nearest integer gives -3.

Answer 7

在c中，右移位运算符保留符号位。因此，在保持符号位的情况下右移位，产生负数，这是两个补码形式。