Does ANSI C support signed / unsigned bit fields?

Question

Does is make sense to qualify bit fields as signed / unsigned?

Answer 1

if a 'bit' is signed, then you have a range of -1, 0, 1, which then becomes a ternary digit. I don't think the standard abbreviation for that would be suitable here, but makes for interesting conversations :)

Answer 2

I don't think Andrew is talking about single-bit bit fields. For example, 4-bit fields: 3 bits of numerical information, one bit for sign. This can entirely make sense, though I admit to not being able to come up with such a scenario off the top of my head.

Update: I'm not saying I can't think of a use for multi-bit bit fields (having used them all the time back in 2400bps modem days to compress data as much as possible for transmission), but I can't think of a use for signed bit fields, especially not a quaint, obvious one that would be an "aha" moment for readers.

Answer 3

Yes, it can. C bit-fields are essentially just limited-range integers. Frequently hardware interfaces pack bits together in such away that some control can go from, say, -8 to 7, in which case you do want a signed bit-field, or from 0 to 15, in which case you want an unsigned bit-field.

Answer 4

The relevant portion of the standard (ISO/IEC 9899:1999) is 6.7.2.1 #4:

A bit-field shall have a type that is a qualified or unqualified version of _Bool, signed int, unsigned int, or some other implementation-defined type.

Answer 5

Yes. An example from here:

struct {
  /* field 4 bits wide */
  unsigned field1 :4;
  /*
   * unnamed 3 bit field
   * unnamed fields allow for padding
   */
  unsigned        :3;
  /*
   * one-bit field
   * can only be 0 or -1 in two's complement!
   */
  signed field2   :1;
  /* align next field on a storage unit */
  unsigned        :0;
  unsigned field3 :6;
}full_of_fields;

Only you know if it makes sense in your projects; typically, it does for fields with more than one bit, if the field can meaningfully be negative.

Answer 6

According to this reference, it's possible:
http://publib.boulder.ibm.com/infocenter/macxhelp/v6v81/index.jsp?topic=/com.ibm.vacpp6m.doc/language/ref/clrc03defbitf.htm

Answer 7

It's very important to qualify your variables as signed or unsigned. The compiler needs to know how to treat your variables during comparisons and casting. Examine the output of this code:

#include <stdio.h>

typedef struct
{
 signed s : 1;
 unsigned u : 1;
} BitStruct;

int main(void)
{
 BitStruct x;

 x.s = 1;
 x.u = 1;

 printf("s: %d \t u: %d\r\n", x.s, x.u);
 printf("s>0: %d \t u>0: %d\r\n", x.s > 0, x.u > 0);

 return 0;
}

Output:

s: -1   u: 1
s>0: 0   u>0: 1

The compiler stores the variable using a single bit, 1 or 0. For signed variables, the most significant bit determines the sign (high is treated negative). Thus, the signed variable, while it gets stored as 1 in binary, it gets interpreted as negative one.

Expanding on this topic, an unsigned two bit number has a range of 0 to 3, while a signed two bit number has a range of -2 to 1.

Answer 8

Bit masking signed types varies from platform hardware to platform hardware due to how it may deal with an overflow from a shift etc.

Any half good QA tool will warn knowingly of such usage.