c - Bitwise shift operators on signed types

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.