Defect Report #122

Submission Date: 03 Dec 93
Submittor: WG14
Source: Ron Guilmette
ANSI/ISO C Defect Report #rfg29:
Subject: Conversion/widening of bit-fields.
Must the following program print 1 or 0?
#include <stdio.h>

struct S { unsigned bit:1; } object = { 1 };

int main ()
printf ("%d\n", ((object.bit - 2) < 0));
return 0;

(At least one existing implementations prints 1 while another prints 0.)
A char, a short int, or an int bit-field, or their signed or unsigned varieties, or an enumeration type, 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 key phrase here is ``the original type.''
In effect, I am asking if the type of a bit-field is totally independent from its width for the purposes of the above rule.
If the answer to that question is ``yes,'' then the value of object.bit must be considered to be an unsigned int (with a value of 1U). In that case, the value 2 used in the above example must also be converted to type unsigned int and then the subtraction should be carried out on the two unsigned int values. The subtraction should then itself yield a value of type unsigned int which is itself (by definition) >= 0, so it would seem that the C Standard requires the above program to print 0.
Is that correct? If so, perhaps the wording of the above paragraph needs to be improved so as to make the correct interpretation of these rules more apparent to implementors.
See Defect Report #015. ``The original type'' applies to both width and signedness. object.bit promotes to int, and the program prints 1.
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