qemu观察arm linux启动过程01 2017-01-18

文来自本人的旧博客 http://blog.163.com/awaken_ing/blog/static/12061319720157611510386

1. 方法

qemu-system-arm -M versatilepb -m 128M -kernel /opt/qemu_arm/linux-2.6.35.7/arch/arm/boot/compressed/vmlinux -S -s

需要留意, 这里-kernel的参数不是/opt/qemu_arm/linux-2.6.35.7/vmlinux

arm-none-linux-gnueabi-gdb -tui -q /opt/qemu_arm/linux-2.6.35.7/arch/arm/boot/compressed/vmlinux
Reading symbols from /opt/qemu_arm/linux-2.6.35.7/arch/arm/boot/compressed/vmlinux...done.
(gdb) target extended-remote localhost:1234
Remote debugging using localhost:1234
warning: Source file is more recent than executable.
start () at arch/arm/boot/compressed/head.S:115
(gdb) b __setup_mmu
Breakpoint 1 at 0x200: file arch/arm/boot/compressed/head.S, line 409.
(gdb) c
Continuing.
Breakpoint 1, __setup_mmu () at arch/arm/boot/compressed/head.S:409
(gdb) i r r3 r4
r3 0x4000 16384
r4 0x8000 32768

qemu观察arm linux启动过程02 2017-01-18

文来自本人的旧博客 http://blog.163.com/awaken_ing/blog/static/12061319720157119242210

1. _createpage_tables

pgtbl r4 @ page table address

其中pgtbl为宏

.macro pgtbl, rd
ldr \rd, =(KERNEL_RAM_PADDR - 0x4000)
.endm

这里KERNELRAMPADDR和zreladdr的值相等.

1.1 identity mapping

/*
* Create identity mapping for first MB of kernel to
* cater for the MMU enable. This identity mapping
* will be removed by paging_init(). We use our current program
* counter to determine corresponding section base address.
*/
mov r6, pc
mov r6, r6, lsr #20 @ start of kernel section
orr r3, r7, r6, lsl #20 @ flags + kernel base
str r3, [r4, r6, lsl #2] @ identity mapping
@ 和arch\arm\boot\compressed\head.S类似
@ r3内容:
@ 31       20       19                 0  
@  PC高12bit       __cpu_mm_mmu_flags
@存放的地址:
@   31          14           13            2                1 0  
@  页表基址(0x4000)         页表index(pc高12bit)              00

arm linux多核同时获取读写锁, 会死循环吗, 还是都得到写锁 2017-01-18

文来自本人的旧博客: http://blog.163.com/awaken_ing/blog/static/120613197201631133058159

本篇只提问, 不解答, 祝开心.

如果读写锁是unlocked, 然后cpu 0和cpu 1都想获取写锁, 如表格所示, CPU 0在t2时刻执行的是wfene, 这时, CPU 1执行的是ldrex, 然后是否会有cpu能够获取到锁? 源码来自 linux-3.10.86/arch/arm/include/asm/spinlock.h archwritelock()

时间	CPU0	时间	CPU1
t0/t6	1: ldrex %0, [%1]
t1/t7	teq %0, #0
t2	WFE("ne")	t2	1: ldrex %0, [%1]
t3	strexeq %0, %2, [%1]	t3	teq %0, #0
t4	teq %0, #0	t4	WFE("ne")
t5	bne 1b	t5	strexeq %0, %2, [%1]
t6		t6	teq %0, #0
t7		t7	bne 1b

中断, softirq, tasklet的区别 2017-01-18

1. 中断流程

假定 CONFIGMULTIIRQHANDLER=y, handlearchirq 为 gichandle_irq.
linux-3.10.86/arch/arm/kernel/setup.c

__irq_svc @entry-armv.S
|   |--svc_entry    保存spsr_<exception> 到r5
|   |--irq_handler
|   |   |--handle_arch_irq => gic_handle_irq
|   |   |   |--irq_find_mapping  完成 an hw irq number到a linux irq的转换
|   |   |   |--handle_IRQ(irqnr, regs) if not IPI 
|   |   |   |       |--generic_handle_irq  -> generic_handle_irq_desc -> handle_fasteoi_irq
|   |   |   |       |   |--raw_spin_lock(&desc->lock);
|   |   |   |       |   |--handle_irq_event
|   |   |   |       |   |   |--raw_spin_unlock(&desc->lock);
|   |   |   |       |   |   |--handle_irq_event_percpu -> action->handler
|   |   |   |       |   |   |--raw_spin_lock(&desc->lock);
|   |   |   |       |   |--chip->irq_eoi
|   |   |   |       |   |--raw_spin_unlock(&desc->lock);
|   |   |   |       |--irq_exit();  这里中断还是禁用的吗? 答:是的, ifndef __ARCH_IRQ_EXIT_IRQS_DISABLED
|   |   |   |       |   |--WARN_ON_ONCE(!irqs_disabled());
|   |   |   |       |   |--if( ... ) invoke_softirq() 
|   |   |   |       |   |    |--do_softirq -> __do_softirq
|   |   |   |       |   |    |    |--local_irq_enable   //修改CPSR I bit, 打开中断
|   |   |   |       |   |    |    |--h->action(h)
|   |   |   |       |   |    |    |--local_irq_disable  //这里为何又禁止中断???
|   |   |   |       |--set_irq_regs(old_regs)  这里开启中断? 答:不是.
|   |--svc_exit r5, irq = 1  @entry-header.S

SLUB 02:frozen 2017-01-14

本文假定没有打开CONFIGSLUBDEBUG, 没有配置CONFIG_NUMA.

1. 什么是frozen

linux-3.10.86/mm/slub.c

 *   If a slab is frozen then it is exempt(免除) from list management. It is not
 *   on any list. It is not on any list. The processor that froze the slab is the one who can
 *   perform list operations on the page. Other processors may put objects
 *   onto the freelist but the processor that froze the slab is the only
 *   one that can retrieve the objects from the page's freelist.

cpu01的c->page是frozen, 那么cpu01可以从该page中取出或放回obj, cpu02不能从该page中取obj, 只能把obj还给该page.

2. cpu partial上的page都是frozen状态

linux-3.10.86/include/linux/slub_def.h

struct kmem_cache_cpu {
...
    struct page *partial;   /* Partially allocated frozen slabs */
    ...
};