Issue 0081: What is the result of the left shift operator E1 < E2, when E1 is signed?

This issue has been automatically converted from the original issue lists and some formatting may not have been preserved.

Authors: Clive Feather, WG14
Date: 1993-12-03
Submitted against: C90
Status: Closed
Converted from: dr.htm, dr_081.html

Item 18 - left shift operator

The result of the left shift operator E1 < E2, when E1 is signed, is defined (subclause 6.3.7) as E1 left-shifted by E2 bits, with vacated bits filled with zeros. But what exactly does this mean?

The C Standard defines a bit (subclause 3.3) only as a unit of data storage. Bits are related to the value of an object only in subclause 6.1.2.5, which specifies the representation of certain types. It may therefore be claimed that the left shift operator must act on representations, which are of fixed length. In this interpretation, the left E2 bits (including the sign bit) are lost, as they would be if E1 was unsigned; the sign bit of the result is taken from a bit in E1, E2 places to the right of the sign bit and, provided that the resultant bit pattern actually represents a value of the result type, an exception is impossible.

On the other hand, it may also be claimed that the whole of subclause 6.3 specifies the meaning of operations in abstract mathematical terms, subject to the general note about exceptions. In this view, the bit sequence representing the non-sign part of a signed integer is converted by the shift operation to a bit sequence of indefinite length, and, to avoid an exception due to overflow, this bit sequence must fit back in the non-sign part without the loss at the left of anything but copies of the sign bit.

a. Which of these two views is correct?

b. If the answer to (a) is the first view, does undefined behavior occur if the resulting bit pattern is not the representation of an integer?

The following questions apply only if the answer to (a) is that the second view is correct.

c. If E1 is positive, and E1 times 2 to the power E2 is less than or equal to INT_MAX (or LONG_MAX), is the result always E1 times 2 to the power E2?

d. Under what circumstances is the result undefined?

Comment from WG14 on 1997-09-23:

Response

Subclause 6.3 states that the binary operator < , among others, has implementation-defined aspects for signed types. Therefore, the answer to “What does it mean to left shift a signed value?” is that it is implementation-defined.