MIT 6.828 Lab1 exercise10

exercise 1.10

要求：在obj/kern/kernel.asm中找到test_backtrace函数，设置一个断点，检查内核启动后每次该函数被调用时发生了什么。多少个32位字在每次嵌套调用test_backtrace时被压入栈？这些被压栈的字是什么？

---

解答：

查询被调用过程，最后指向了一个C文件，lab/kern/init.c 其中有该函数的代码。

test_backtrace(int x)
{
	cprintf("entering test_backtrace %d\n", x);
	if (x > 0)
		test_backtrace(x-1);
	else
		mon_backtrace(0, 0, 0);
	cprintf("leaving test_backtrace %d\n", x);
}

显然是个递归调用，循环输出结果如下（推知）

//假设x=3
entering test_backtrace 3
entering test_backtrace 2
entering test_backtrace 1
entering test_backtrace 0
mon_backtrace(0, 0, 0)
leaving test_backtrace 0
leaving test_backtrace 1
leaving test_backtrace 2
leaving test_backtrace 3

回到kernel.asm，line 75就是这个函数的汇编代码,f复制过来如下。能看到把ebp、ebx、eax压栈，ebx压栈两次，则共压栈4个32位字。

// Test the stack backtrace function (lab 1 only)
void
test_backtrace(int x)
{
f0100040:	55                   	push   %ebp
f0100041:	89 e5                	mov    %esp,%ebp
f0100043:	53                   	push   %ebx
f0100044:	83 ec 0c             	sub    $0xc,%esp
f0100047:	8b 5d 08             	mov    0x8(%ebp),%ebx
	cprintf("entering test_backtrace %d\n", x);
f010004a:	53                   	push   %ebx
f010004b:	68 80 18 10 f0       	push   $0xf0101880
f0100050:	e8 a3 08 00 00       	call   f01008f8 <cprintf>
	if (x > 0)
f0100055:	83 c4 10             	add    $0x10,%esp
f0100058:	85 db                	test   %ebx,%ebx
f010005a:	7e 11                	jle    f010006d <test_backtrace+0x2d>
		test_backtrace(x-1);
f010005c:	83 ec 0c             	sub    $0xc,%esp
f010005f:	8d 43 ff             	lea    -0x1(%ebx),%eax
f0100062:	50                   	push   %eax
f0100063:	e8 d8 ff ff ff       	call   f0100040 <test_backtrace>
f0100068:	83 c4 10             	add    $0x10,%esp
f010006b:	eb 11                	jmp    f010007e <test_backtrace+0x3e>
	else
		mon_backtrace(0, 0, 0);
f010006d:	83 ec 04             	sub    $0x4,%esp
f0100070:	6a 00                	push   $0x0
f0100072:	6a 00                	push   $0x0
f0100074:	6a 00                	push   $0x0
f0100076:	e8 f3 06 00 00       	call   f010076e <mon_backtrace>
f010007b:	83 c4 10             	add    $0x10,%esp
	cprintf("leaving test_backtrace %d\n", x);
f010007e:	83 ec 08             	sub    $0x8,%esp
f0100081:	53                   	push   %ebx
f0100082:	68 9c 18 10 f0       	push   $0xf010189c
f0100087:	e8 6c 08 00 00       	call   f01008f8 <cprintf>
}
f010008c:	83 c4 10             	add    $0x10,%esp
f010008f:	8b 5d fc             	mov    -0x4(%ebp),%ebx
f0100092:	c9                   	leave  
f0100093:	c3                   	ret