【MIPS CPU 体系结构概述1】连载2:

3. 寄存器约定

对于在一个CPU上进行开发,掌握其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 $fp 帧指针(Frame Pointer)
(fp在现代编译器中基本上不用了,而是作为一个$t8来使用。) $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 为编译器会处理。但对于操作系统核心或驱动程序开发人员就必须清楚。
一般来讲,你通过objdump -d可以清醒的看到寄存器的用法。 下面通过笔者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: Huailin Chen
* 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。

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

~/huailinchen: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 optimization.
* With mips architecture, the instruction after jump
* will also be fetched into the pipline and get
* executed. 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

希望读者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 她们保存起来,然后再恢复,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。

为了使得读者更加理解寄存器的用法,下面笔者特意提出的一个问题。
*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?

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