Issue 0015: How does an unsigned plain bitfield promote?

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

Authors: Craig Blitz, WG14
Date: 1992-12-10
Reference document: X3J11/90-051
Submitted against: C90
Status: Closed
Cross-references: 0122
Converted from: dr.htm, dr_015.html

This question concerns the promoted type of plain int bit-fields with length equal to the size of an object of type int. I am interested in implementations that have chosen not to regard the high-order bit as a sign bit.

The question is: What is the promoted type of such an object?

Subclause 6.5.2.1 states:

A bit-field shall have a type that is ... int, unsigned int, or signed int.

The intent of this, I believe, is that the type of a plain int bit-field is int.

Subclause 6.2.1.1 states:

A char, a short int, or an int b it-field, or their signed or unsigned varieties, ... may be used in an expression wherever an int or unsigned int may be used. If an int can represent all values of the original type, the value is converted to an int; otherwise it is converted to an unsigned int...

The integral promotions preserve value including sign.

Tracing this through, then, the type of any promoted plain int bit-field is int, since int can hold all the values of the original type, which is int. However, not all values of the bit-field, which may be regarded as non-negative, can be represented by an int. By value-preserving promotion rules, I would expect the type of the promoted bit-field to be unsigned int.

Can you clarify this?

Comment from WG14 on 1997-09-23:

Response

As described in subclause 6.2.1.1, bit-fields that are being treated as unsigned will promote according to the same rules as other unsigned types: if the width is less than int, and int can hold all the values, then the promotion is to int. Otherwise, promotion is to unsigned int.