/* Copyright 1997 Acorn Computers Ltd
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
/***************************************************/
/* File   : Bitfields.h                            */
/*                                                 */
/* Purpose: A few macros and definitions to handle */
/*          large bitfields, created originally    */
/*          for recording used font handles.       */
/*                                                 */
/* Author : A.D.Hodgkinson, adapted from the       */
/*          TCPIPLibs source sys/types.h           */
/*                                                 */
/* History: 10-Mar-97: Created.                    */
/*          16-Apr-97: Modified to vary the size   */
/*                     of the bitfield according   */
/*                     to Limits_OS_FontHandles.   */
/***************************************************/

#include "Limits.h"

#define Bits_In_A_Byte 8
#define Bitfield_Size  Limits_OS_FontHandles

typedef long bitfield_mask;

#define Number_Of_Bits     (sizeof(bitfield_mask) * Bits_In_A_Byte)
#define How_Many_Bits(x,y) (((x) + ((y) - 1)) / (y))

typedef struct bitfield_set
{
  bitfield_mask bitfield_bits[How_Many_Bits(Bitfield_Size, Number_Of_Bits)];

} bitfield_set;

#define Bitfield_Set_Bit(p,n)   ((p)->bitfield_bits[(n) / Number_Of_Bits] |=  (1l << ((n) % Number_Of_Bits)))
#define Bitfield_Clear_Bit(p,n) ((p)->bitfield_bits[(n) / Number_Of_Bits] &= ~(1l << ((n) % Number_Of_Bits)))
#define Bitfield_Is_Set(p,n)    ((p)->bitfield_bits[(n) / Number_Of_Bits] &   (1l << ((n) % Number_Of_Bits)))