MIPS体系结构剖析，编程与实践【2】

MIPS体系结构剖析，编程与实践【2】

第二章 寄存器约定

寄存器约定

对于在一个CPU上进行开发，掌握其工作的CPU的寄存器约定是非常重要的。

MIPS体系结构提供了32个GPR(GENERAL PURPOSE REGISTER)。这32个寄存器的用法大致如下：

REGISTER NAME USAGE
$0 $zero 常量0(constant value 0)
$2-$3 $v0-$v1 函数调用返回值(values for results and expression evaluation)
$4-$7 $a0-$a3 函数调用参数(arguments)

$8-$15 $t0-$t7 暂时的(或随便用的)
$16-$23 $s0-$s7 保存的(或如果用，需要SAVE/RESTORE的)(saved)
$24-$25 $t8-$t9 暂时的(或随便用的)
$28 $gp 全局指针(Global Pointer)
$29 $sp 堆栈指针(Stack Pointer)
$30 $fp 帧指针(Frame Pointer)
(BNN：fp is stale acutally, and can be simply used as $t8)
$31 $ra 返回地址(return address)

对一个CPU的寄存器约定的正确用法是非常重要的。当然对C语言开发者不需要关心，因为COMPILER会TAKE CARE。但对于KERNEL的开发或DRIVER开发的人就**必须**清楚。

一般来讲，你通过objdump -d可以清醒的看到寄存器的用法。

下面通过我刚才写的一个简单例子来讲解：

~/ vi Hello.c
"Hello.c" [New file]
/* Example to illustrate mips register convention
* -Author: BNN
* 11/29/2001
*/

int addFunc(int,int);
int subFunc(int);

void main()
{

int x,y,z;
x= 1;
y=2;
z = addFunc(x,y);
}


int addFunc(int x,int y)
{
int value1 = 5;
int value2;

value2 = subFunc(value1);
return (x+y+value2);

}

int subFunc(int value)
{
return value--;
}

上面是一个C程序，main()函数调用一个加法的子函数。让我们来看看编译器是如何产生代码的。

~/bnn:74> /bin/mips-elf-gcc -c Hello.o Hello.c -mips3 -mcpu=r4000 -mgp32 -mfp32 -O1

~/bnn:75> /bin/mips64-elf-objdump -d Hello.o
Hello.o: file format elf32-bigmips
Disassembly of section .text:

/* main Function */
0000000000000000 :
/*create a stack frame by moving the stack pointer 8
*bytes down and meantime update the sp value
*/
0: 27bdfff8 addiu $sp,$sp,-8
/* Save the return address to the current sp position.*/
4: afbf0000 sw $ra,0($sp)
8: 0c000000 jal 0
/* nop is for the delay slot */
c: 00000000 nop
/* Fill the argument a0 with the value 1 */
10: 24040001 li $a0,1
/* Jump the addFunc */
14: 0c00000a jal 28
/* NOTE HERE: Why we fill the second argument
*behind the addFunc function call?
* This is all about the "-O1" compilation optimizaiton.
* With mips architecture, the instruciton after jump
* will also be fetched into the pipline and get
* exectuted. Therefore, we can promise that the
* second argument will be filled with the value of
* integer 2.
*/
18: 24050002 li $a1,2
/*Load the return address from the stack pointer
* Note here that the result v0 contains the result of
* addFunc function call
*/
1c: 8fbf0000 lw $ra,0($sp)
/* Return */
20: 03e00008 jr $ra
/* Restore the stack frame */
24: 27bd0008 addiu $sp,$sp,8

/* addFunc Function */
0000000000000028 :
/* Create a stack frame by allocating 16 bytes or 4
* words size
*/
28: 27bdfff0 addiu $sp,$sp,-16
/* Save the return address into the stack with 8 bytes
* offset. Please note that compiler does not save the
* ra to 0($sp).
*Think of why, in contrast of the previous PowerPC
* EABI convention
*/
2c: afbf0008 sw $ra,8($sp)
/* We save the s1 reg. value into the stack
* because we will use s1 in this function
* Note that the 4,5,6,7($sp) positions will then
* be occupied by this 32 bits size register
*/
30: afb10004 sw $s1,4($sp)
/* Withe same reason, save s0 reg. */
34: afb00000 sw $s0,0($sp)
/* Retrieve the argument 0 into s0 reg. */
38: 0080802d move $s0,$a0
/* Retrieve the argument 1 into s1 reg. */
3c: 00a0882d move $s1,$a1
/* Call the subFunc with a0 with 5 */
40: 0c000019 jal 64
/* In the delay slot, we load the 5 into argument a0 reg
*for subFunc call.
*/
44: 24040005 li $a0,5
/* s0 = s0+s1; note that s0 and s1 holds the values of
* x,y, respectively
*/
48: 02118021 addu $s0,$s0,$s1
/* v0 = s0+v0; v0 holds the return results of subFunc
*call; And we let v0 hold the final results
*/
4c: 02021021 addu $v0,$s0,$v0
/*Retrieve the ra value from stack */
50: 8fbf0008 lw $ra,8($sp)
/*!!!!restore the s1 reg. value */
54: 8fb10004 lw $s1,4($sp)
/*!!!! restore the s0 reg. value */
58: 8fb00000 lw $s0,0($sp)
/* Return back to main func */
5c: 03e00008 jr $ra
/* Update/restore the stack pointer/frame */
60: 27bd0010 addiu $sp,$sp,16

/* subFunc Function */
0000000000000064 :
/* return back to addFunc function */
64: 03e00008 jr $ra
/* Taking advantage of the mips delay slot, filling the
* result reg v0 by simply assigning the v0 as the value
*of a0. This is a bug from my c source
* codes--"value--". I should write my codes
* like "--value", instead.
68: 0080102d move $v0,$a0


希望大家静下心来把上面的代码看懂。一定要注意编译器为什么在使用s0和s1之前要先把她们SAVE起来，然后再RESTORE，虽然在这个例子中虽然main 函数没用s0和s1。


另外的一点是：由于我们加了“-O1”优化，编译器利用了“delay slot"来执行那些必须执行的指令，而不是简单的塞一个”nop"指令在那里。非常的漂亮。

最后，考大家一个问题，为了使得大家更加理解寄存器的用法：

*在写一个核心调度context switch()例程时，我们需要SAVE/RESTORE$t0-$t7吗？如果不，为什么？

*在写一个时钟中断处理例程时，我们需要SAVE/RESTORE$t0-$t7吗？如果是，为什么？