问题描述
我了解到，只要我在MIPS中拥有一个具有四个以上参数的函数，就应该利用堆栈.但是在下面的代码中，将第五个参数保存在sw $t0, 4($sp)并执行jal sad之后，然后在sad函数的开头，我再次调整堆栈指针以保存由$sx使用的$sx寄存器.呼叫者.我在这里做错什么了吗?
I understand that whenever I have a function that has more than four arguments in MIPS I should utilize the stack. However in my code below after saving the fifth argument at sw $t0, 4($sp) and do a jal sad, then right at the beginning of the sad function I adjust the stack pointer again to save the $sx registers that is used by the caller. Am I doing something wrong here?
vbsme: subu $sp, $sp, 8 # create space on the stack pointer
sw $ra, 0($sp) # save return address
li $v0, 0 # reset $v0
li $v1, 0 # reset $v1
li $s0, 1 # i(row) = 1
li $s1, 1 # j(col) = 1
lw $s2, 0($a0) # row size
lw $s3, 4($a0) # col size
mul $s4, $s2, $s3 # row * col
li $s5, 0 # element = 0
loop: bgeq $s5, $s4, exit # if element >= row * col then exit
subi $a3, $s0, 1 # 4th parameter: i-1
subi $t0, $s1, 1
sw $t0, 4($sp) # 5th parameter: j-1
jal sad # calculate the sum of absolute difference using the frame starting from row a3 and col 4($sp)
add $s6, $s0, $s1
andi $s7, $s6, 1
if: bneq $s7, $zero, else
inif: bge $s1, $s2, inelse
addi $s1, $s1, 1
j inif1
inelse: addi $s0, $s0, 2
inif1: subi $s7, $s0, 1
beq $s7, $zero, loop_back
subi $s0, $s0, 1
j loop_back
else: bge $s0, $s2, inelse1
addi $s0, $s0, 1
j inif2
inelse1:addi $s1, $s1, 2
inif2: subi $s7, $s1, 1
beq $s7, $zero, loop_back
subi $s1, $s1, 1
j loop_back
loop_back: addi $s5, $s5, 1
j loop
exit: lw $ra, 0($sp) # restore return address
addi $sp, $sp, 8 # restore stack pointer
jr $ra # return
.globl sad
sad: subu $sp, $sp, 32 # allocate stack space for largest function
sw $s7, 28($sp) # save $s7 value
sw $s6, 24($sp) # save $s6 value
sw $s5, 20($sp) # save $s5 value
sw $s4, 16($sp) # save $s4 value
sw $s3, 12($sp) # save $s3 value
sw $s2, 8($sp) # save $s2 value
sw $s1, 4($sp) # save $s1 value
sw $s0, 0($sp) # save $s0 value
#some code to be filled later
lw $s7, 28($sp) # restore original value of $s7 for caller
lw $s6, 24($sp) # restore original value of $s6 for caller
lw $s5, 20($sp) # restore original value of $s5 for caller
lw $s4, 16($sp) # restore original value of $s4 for caller
lw $s3, 12($sp) # restore original value of $s3 for caller
lw $s2, 8($sp) # restore original value of $s2 for caller
lw $s1, 4($sp) # restore original value of $s1 for caller
lw $s0, 0($sp) # restore original value of $s0 for caller
addiu $sp, $sp, 32 # restore the caller's stack pointer
jr $ra # return to caller's code
推荐答案
这是gcc的工作方式.有关更多信息，您(可以)应阅读Mips ABI.有些事情可能有所不同.
This is how its done by gcc. For more information, you (could) should readthe Mips ABI. Some things may differ.
http://math-atlas.sourceforge.net/devel/assembly/mipsabi32.pdf
按照惯例，第五个参数应放在堆栈的第五个字上.
By convention, the fifth argument should go on the fifth word of the stack.
所以您应该
sad:
sub $sp,$sp,24 #24 byte stack frame
... some code ...
#Convention indicates to store $a0..$a3 in A..D (see below)
sw $a0,0(sp)
sw $a1,4(sp)
sw $a2,8(sp)
sw $a3,12(sp)
#Get the 5th argument
lw $t0,40($sp) #40 : 24 + 16
要将第5个参数存储在堆栈中，您应该了解以下信息:
To store the 5th argument in the stack, you should know this:
如果vbsme将要调用另一个函数，则应保存堆栈的底部4个字，以便被调用方在此处存储参数值.如果传递了四个以上的参数，则应为每个参数保存一个额外的单词.
If vbsme is going to call another function, then the bottom 4 words of the stack should be saved for the callee to store argument values there. If more than 4 arguments are passed, then an additional word should be saved for each argument.
vbsme's stack frame bottom part (Argument building area)
| ... |
---------------
| 5th arg | <---- sw $t5,16($sp)
---------------
| D |
---------------
| C |
---------------
| B |
---------------
| A |
--------------- <--sp (of vbsme stack frame)
此外，$ ra寄存器应该保存在堆栈的顶部，因为它的寄存器是31.
Also, the $ra register should be saved at the top of the stack, since its register 31.
vbsme:
subu $sp, $sp, 20+N # 20: space for 5 arguments,
#N space for other stuff (ra,$tx, etc)
#Set arguments (assumes 5th parameter value is in register $t5)
subi $a3, $s0, 1 # 4th parameter: i-1
sw $t5,16($sp) #
...
.end
响应
Why is it that you do:
lw $t0,40($sp)
to get the 5th argument, why did you add 24 to 16? when you do
sub $sp,$sp,24
don't you already move
the sp 24 place?
是的，$ sp + 24指向调用方堆栈的底部.但是，那不是我提出第五个论点的地方.第五个参数放在调用程序堆栈的第五个单词上，这就是为什么我要添加16.
Yes, $sp + 24 points to the base of the caller´s stack. However, thats not where I placed the fifth argument. The fifth argument is placed on the fifth word of the callers stack, thats why I add 16.
这篇关于具有四个以上参数的MIPS函数调用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！