linux 1.0 内核注解 linux/fs/ext2/bitmap.c
时间：2009-05-24 来源：taozhijiangscu
/********************************************
*Created By: Prometheus
*Date : 2009-5-24
********************************************/
/*
* linux/fs/ext2/bitmap.c
*
* Copyright (C) 1992, 1993, 1994 Remy Card ([email protected])
* Laboratoire MASI - Institut Blaise Pascal
* Universite Pierre et Marie Curie (Paris VI)
*/ #include <linux/fs.h>
#include <linux/ext2_fs.h> //这里总共有16个数，他们是0x00-0x0f中0的数目，用半字节为单位来统计空闲的位图
static int nibblemap[] = {4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0}; unsigned long ext2_count_free (struct buffer_head * map, unsigned int numchars)
{
unsigned int i;
unsigned long sum = 0;
if (!map)
return (0);
for (i = 0; i < numchars; i++)
sum += nibblemap[map->b_data[i] & 0xf] + //低4比特
nibblemap[(map->b_data[i] >> 4) & 0xf]; //高4比特
return (sum);
}
文档地址:http://blogimg.chinaunix.net/blog/upfile2/090524225100.pdf
*Created By: Prometheus
*Date : 2009-5-24
********************************************/
/*
* linux/fs/ext2/bitmap.c
*
* Copyright (C) 1992, 1993, 1994 Remy Card ([email protected])
* Laboratoire MASI - Institut Blaise Pascal
* Universite Pierre et Marie Curie (Paris VI)
*/ #include <linux/fs.h>
#include <linux/ext2_fs.h> //这里总共有16个数，他们是0x00-0x0f中0的数目，用半字节为单位来统计空闲的位图
static int nibblemap[] = {4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0}; unsigned long ext2_count_free (struct buffer_head * map, unsigned int numchars)
{
unsigned int i;
unsigned long sum = 0;
if (!map)
return (0);
for (i = 0; i < numchars; i++)
sum += nibblemap[map->b_data[i] & 0xf] + //低4比特
nibblemap[(map->b_data[i] >> 4) & 0xf]; //高4比特
return (sum);
}
文档地址:http://blogimg.chinaunix.net/blog/upfile2/090524225100.pdf
相关阅读 更多 +
