Does ANSI C support signed / unsigned bit fields?

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.

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.

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: 1s>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.