签名类型的按位移位运算符

Question

I am trying to understand the behavior of bitwise operators on signed and unsigned types. As per the ISO/IEC document, following are my understandings.

Left shift Operator

• The result of E1 << E2 , is E1 left-shifted E2 bit positions

• The vacated bits on the account of left shift will be filled by zeros.

• E1 as signed non-negative: E1 << E2 will result to E1 multiplied by 2 power of E2, if the value is representable by the result type.

• Q1: What about signed negatives?

• Q2: I am not able to understand on what's meant by "reduced modulo" in the following context. "If E1 has an unsigned type, the value of the result is E1 × 2E2 , reduced modulo one more than the maximum value representable in the result type" .

Right Shift Operator

• The result of E1 >> E2 is E1 right-shifted E2 bit positions.

• E1 as signed non-negative/unsigned :The value of the result is the integral part of the quotient of E1 / 2E2

• Q3: For signed negative integers I see, some books defining that the vacant positions will be filled with 1 .Please elaborate more on the use of right shift operator on signed negative int.

Answer 1

Q1: The behaviour of the left shift operator on negative values of signed integer types is undefined, as is the behaviour for positive values of signed integer types when the result E1 * 2^E2 is not representable in the type.

That is explicitly mentioned in section 6.5.7, paragraph 4 and 5 of the standard (n1570 draft):

4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

Q2: The reduction modulo one more than the maximum value representable in the unsigned integer type means that the bits that are shifted out on the left are simply discarded.

Mathematically, if the maximal value of the unsigned type is 2^n - 1 (and it is always of that form), the result of shifting E1 left by E2 bits is the value V in the range from 0 to 2^n - 1 such that the difference

(E1 * 2^E2 - V)

is divisible by 2^n , that is, it's the remainder you get when dividing E1 * 2^E2 by 2^n .

Q3: The behaviour upon shifting right negative values of signed integer types is implementation-defined. The most common behaviour (at least on two's complement machines) is an arithmetic shift, that is, the result is the quotient rounded down (towards negative infinity).

5 The result 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 nonnegative value, the value of the result is the integral part of the quotient of E1 / 2^E2 . If E1 has a signed type and a negative value, the resulting value is implementation-defined.

Answer 2

• Re: Q1
  If E1 is negative, the behavior is undefined.

• Re: Q2
  Unsigned arithmetic is "cyclic", that is, it wraps around so UINT_MAX + 1 is 0 again. It's as if every calculation was done modulo UINT_MAX+1. Another way to think about it is that excess bits that don't fit in on the left are simply dropped.

• Re: Q3
  If E1 is negative, the result is implementation-defined. That is, it depends on your machine/compiler/options, but the behavior has to be documented ("defined") somewhere, ususally the compiler manual. Two popular choices are to fill the incoming bits on the left with 1 (arithmetic shift) or with 0 (logical shift).

Answer 3

If you really want to understand the bitwise shift operators. Look at these simple rules:

1) In left shift, E1 << E2, all the vacant bits on right side will be filled by zeroes, does not matter if the number is signed or unsigned, always zeroes will be shifted in.

2) In left shift, E1 >> E2, all the vacant bits on left side, will be 0 if number is positive, and will be 1 if number is negative.Keep in mind that an unsigned number is never negative. Also some implementations might also fill them with 0 on some machines even if number is negative, so never rely on this.

All other scenarios could be explained by these two simple rules. Now if you want to know the value of result after shifting , just write the bit representation of number and shift manually on paper and enter bits at vacant spaces using these two rules. Then you would be able to understand better how bit shifting works.

For example lets take int i = 7;
i<<2
now i = 0000 0000 0000 0000 0000 0000 0000 0111
perform two left shits according to rule 1 would be:

0000 0000 0000 0000 0000 0000 0001 1100

which will give it value of 28 ( E1 * 2E2 = 7 *2*2 = 28), which is same as representated by bit pattern.

Now let me explain the "reduced modulo" thing, well if you are shifting a big number, than the bits on left hand side will be lost, "reduced modulo" compensates for it, So if your resultant value if greater than the maximum value that the data type could hold, the bits on left will be lost , and then: result = (E1*2*E2) % ( maximumValue + 1).

For various other cases, just keep the above rules in mind , and you are good :)

Answer 4

Q2：“降值模数X”在数学中表示“值mod X”，可以在C中写为“value％X”。这部分只是解释整数溢出是如何工作的。