2.4.22 i386 VM 分享 MAXMEM

最新推荐文章于 2025-12-19 15:16:31 发布

原创最新推荐文章于 2025-12-19 15:16:31 发布 · 397 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#测试 #include #system #x86

本文深入探讨了Linux内核中用于管理内存映射和位操作的机制，包括内存区域的划分、内存映射的打印及内核中使用的原子操作。通过解析关键代码段，解释了如何高效地管理和利用内存资源，以及如何实现原子操作以确保数据的一致性和可靠性。

/* include\asm-i386\page.h */

#define MAXMEM ((unsigned long)(-PAGE_OFFSET-VMALLOC_RESERVE))

#define PAGE_OFFSET ((unsigned long)__PAGE_OFFSET)

#define VMALLOC_RESERVE ((unsigned long)__VMALLOC_RESERVE)

#define __PAGE_OFFSET (0xC0000000)

#define __VMALLOC_RESERVE (128 << 20)

VMALLOC_RESERVE : 0x08000000

PAGE_OFFSET : 0xc0000000

VMALLOC_RESERVE + PAGE_OFFSET = 0xc8000000 :-) foo

再求个负数嘛(等价于取反加1) 0x38000000 (3*256 + 8 * 16 ) M = 896M zone_normal

===========================================================================================

static void __init print_memory_map(char *who)
{
int i;

for (i = 0; i < e820.nr_map; i++) {
  printk(" %s: %016Lx - %016Lx ", who,
   e820.map[i].addr,
   e820.map[i].addr + e820.map[i].size);
  switch (e820.map[i].type) {
  case E820_RAM: printk("(usable)\n");
    break;
  case E820_RESERVED:
    printk("(reserved)\n");
    break;
  case E820_ACPI:
    printk("(ACPI data)\n");
    break;
  case E820_NVS:
    printk("(ACPI NVS)\n");
    break;
  default: printk("type %lu\n", e820.map[i].type);
    break;
  }
}
}

switch (e820.map[i].type) {
  case E820_RAM: res->name = "System RAM"; break;
  case E820_ACPI: res->name = "ACPI Tables"; break;
  case E820_NVS: res->name = "ACPI Non-volatile Storage"; break;
  default: res->name = "reserved";
  }

==========================================================

test_and_set_bit

@nr: Bit to set
@addr: Address to count from
注意 .This operation is atomic and cannot be reordered.It may be reordered on other architectures than x86.It also implies a memory barrier.
====> 所以要用锁LOCK_PREFIX

static inline int test_and_set_bit(int nr, volatile unsigned long * addr)
{
int oldbit;

__asm__ __volatile__( LOCK_PREFIX
"btsl %2,%1\n\tsbbl %0,%0"
:"=r" (oldbit),"=m" (ADDR)
:"Ir" (nr)
:"memory");
return oldbit;
}
分析：
bts 测试位并置位，被测试位送CF并且被测试位置1 与btr相对。

sbbl %0,%0 将 ”-CF“ 给%0