Issue 0335: _Bool bit-fields

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

Authors: WG 14, Fred Tydeman (USA)
Date: 2006-12-12
Reference document: ISO/IEC WG14 N1204, ISO/IEC WG14 DR 262, ISO/IEC WG14 DR 315
Submitted against: C99
Status: Closed
Cross-references: 0262, 0315
Converted from: summary-c99.htm, dr_335.htm

Summary

What are the constraints on and semantics of _Bool bit-fields?

  #include <stdbool.h>
  struct bits {
      _Bool    bbf1 : 1;        /* unsigned 1-bit _Bool bit-field */
      _Bool    bbf3 : 3;        /* unsigned 3-bit _Bool bit-field */
  } bits;
  int main(void){
    bits.bbf1 = true;              /* the value 1u */
    bits.bbf1 = ~ bits.bbf1;       /* undefined? 0u? 1u? */
    bits.bbf3 = true;              /* the value 1u */
    bits.bbf3 = ~ bits.bbf3;       /* undefined? 0u? 1u? 6u? */
    return 0;
  }

What is the maximum width of a _Bool bit-field allowed that does not cause a constraint violation? 1? CHAR_BIT? Something else? Is bbf3 a constraint violation?

DR 262 changed 6.7.2.1#3 to require a constraint violation if a bit-field width is too large.

6.2.5#2 says _Bool can hold the values 0 and 1.

6.2.6.2#6 discusses width.

I see nothing that says the width of a _Bool is 1.

6.7.2.1#9 has: If the value 0 or 1 is stored into a nonzero-width bit-field of type _Bool, the value of the bit-field shall compare equal to the value stored.

So, if a value other than 0 or 1 is stored into a nonzero-width bit-field of type _Bool, is that undefined?

The first assignment gives the bit-field the value 1u. The ~ of that value yields ...1111110u. When that value is then stored into the _Bool bit-field, what value is stored?

• 0u since it is a 1-bit unsigned int field.
• 1u since it is a _Bool (and the value is non-zero).
• 6u since it is a 3-bit unsigned int field.
• Undefined.

6.3.1.2 Boolean type has: When any scalar value is converted to _Bool, the result is 0 if the value compares equal to 0; otherwise, the result is 1.

6.7.2.1#9 has: A bit-field is interpreted as a signed or unsigned integer type consisting of the specified number of bits.

So, does a _Bool bit-field have the semantics of a _Bool (as per 6.3.1.2) or of an unsigned integer type (as per 6.7.2.1)? DR 315 might be relevant.

Suggested Technical Corrigendum

Comment from WG14 on 2008-07-21:

Proposed Technical Corrigendum

Committee Discussion (for history only)

Spring 2007

The width of a _Bool bit-field is at most the implementation defined width of the type _Bool. A _Bool bit-field has the semantics of a _Bool (and not an unsigned int).

Spring 2008

6.7.2.1 paragraph 3 covers the above Committee discussion. (9899:1999 + TC1 + TC 2 + TC3)

The expression that specifies the width of a bit-field shall be an integer constant expression with a nonnegative value that does not exceed the width of an object of the type that would be specified were the colon and expression omitted.

Therefore the width of a _Bool bit-field is at most the implementation-defined width of the type _Bool.

Committee Response

6.2.5 paragraph 6 states that

The type _Bool and the unsigned integer types that correspond to the standard signed integer types are the standard unsigned integer types.

In other words, _Bool is one of the unsigned integer types whether it is used in a bit-field or not. 6.3.1.2p1 explicitly defines the semantics of _Bool, which are different from other unsigned integer types.

A _Bool bit-field has the semantics of a _Bool (and not unsigned int).

6.7.2.1 paragraph 3 (9899:1999 + TC1 + TC 2 + TC3) states that

The expression that specifies the width of a bit-field shall be an integer constant expression with a nonnegative value that does not exceed the width of an object of the type that would be specified were the colon and expression omitted.

The width of a _Bool bit-field is at most the implementation-defined width of the type _Bool.