:
/*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？