ARM 汇编器常规冒泡排序 [英] ARM Assembler Regular Bubble Sort
问题描述
我尝试根据给定的 C 冒泡排序代码在 ARM 汇编程序中编写一个简单的冒泡排序:
I tried to write a simple bubblesort in ARM Assembler on a basis of a given C bubblesort code:
bubbleSort (int A[], int n)
{
int temp;
int i;
do {
swapped = false;
n = n-1;
for (i=0; i < n; i++) {
if ( A[i] > A[i+1] ) {
temp = A[i];
A[i] = A[i+1];
A[i+1] = temp;
swapped = true;
}
}
} while (swapped == true);
}
这是我目前编写的汇编代码:
And here is my assembler code I wrote so far:
bubblesort:
@ maybe r0, r1 for table and counter (rest is based on that)
@ r8 = Buffer for Value
@ r4 = Counter for loop1
@ r5 = Register for Value
@ r7 = Register for Value
@ r1 = Size of Array
@ r0 = Address of the 1. Element
@ r10 = "swapped" register
loop:
mov r10, #0
sub r1, #1 @ Decrement Counter for loop
loop1:
ldr r5, [r0]
ldr r7, [r0, #4]
cmp r5, r7
movgt r8, r7
movgt r7, r5
movgt r5, r8
strgt r5, [r0]
strgt r7, [r0, #4]
movgt r10, #1
cmp r4, r1 @ compare r4 and r1
add r4, #1 @ Increment counter for loop1
blt loop1 @ is r4 < r1 ? yes = go to loop1
cmp r10, #1
beq loop
我的目标是对整数序列或数组"进行排序.给定参数是第一个值的地址 (r0) 和表大小 (r1).将内存中的值加载到寄存器中，比较它们，然后交换寄存器中的值并将它们存储到内存中的正确位置.
My goal is to sort a sequence of ints or an "array". Given parameters are the address of the first value (r0) and the table size (r1). Load the values from the memory into registers, comparing them, then swap the values in the registers and store them at the right position into the memory.
之前我更改了运行调试器的代码，寄存器中的交换"部分似乎工作正常.但我不太确定我在那里做的循环逻辑.我的问题是，当我查看内存时，第一个值被正确排序，然后它停止排序/忽略其余部分.
Before I changed my code I ran the debugger, the "swapping" part in the registers seems to work fine. But im not too sure about the loop-logic I did there. My problem is, when I take a look into the memory, the first value is correctly sorted and then it just stops sorting / ignores the rest.
现在我得到的只是一个无限循环.
Now all I get is an endless loop.
我将非常感谢任何给定的帮助.
I will greatly appreciate any given help.
另外，如果有什么我忘了提，请告诉我，我会提供的.
Also if there is anything I forgot mentioning, just let me know and I will provide it.
提前致谢，
德特
推荐答案
@ ---- Added ----
@ bubblesort.s
@
@ pi@RPi3:~ $ as -o bubblesort.o bubblesort.s
@ pi@RPi3:~ $ gcc -o bubblesort bubblesort.o
@ pi@RPi3:~ $ ./bubblesort; echo $?
.data
table:
.word 9, -8, -7, 6, -5, 4, -3, 2, -1, 0, 127, -128
aft_table:
.balign 4
n:
.word (aft_table - table) >>2
.text
.global main
main:
push {r4-r10, lr}
ldr r1, =n
ldr r1, [r1] @ len(table) in r1
@ ---------------
bubblesort:
@ maybe r0, r1 for table and counter (rest is based on that)
@ r8 = Buffer for Value
@ r4 = Counter for loop1
@ r5 = Register for Value
@ r7 = Register for Value
@ r1 = Size of Array
@ r0 = Address of the 1. Element
@ r10 = "swapped" register
loop:
mov r10, #0
sub r1, #1 @ Decrement Counter for loop
@ ---- Added ----
ldr r0, =table
mov r4, #1 @ i = 1 i could = 0, if instructions
@ cmp r4, r1 @ compare r4 and r1
@ add r4, #1 @ Increment counter for loop1
@ were reversed
@ ---------------
loop1:
ldr r5, [r0]
ldr r7, [r0, #4]
cmp r5, r7
movgt r8, r7
movgt r7, r5
movgt r5, r8
strgt r5, [r0]
strgt r7, [r0, #4]
movgt r10, #1
@ ---- Added ----
add r0, #4 @ Index table to next position
@ ---------------
cmp r4, r1 @ compare r4 and r1
add r4, #1 @ Increment counter for loop1
blt loop1 @ is r4 < r1 ? yes = go to loop1
cmp r10, #1
beq loop
@ ---- Added ----
exit:
pop {r4-r10, lr}
mov r0, #0 @ good return code
bx lr @ Exit with gcc linker
@ ---------------
你的代码逻辑很好.我添加了几个部分来编译和运行它，这在您的代码中进行了标记.需要注意两点，mov r4,#1 开始计数，add r0,#4 将表指针移动到下一个位置.
Your code logic is good. I have added several sections to compile and run it, which is marked in your code. Two things to note, mov r4, #1 to start the count, and add r0, #4 to move the table pointer to the next position.
我还将包含我的调试脚本，这些脚本有助于发现错误并帮助理解交换开关(通过将 -128 更改为 128，减少循环次数).
I will also include my debugging scripts that helped with finding errors and help to understand the swap switch (by changing the -128 to 128, less passes through the loops).
@ ---- Added ----
@ bubblesort2.s
@
@ pi@RPi3:~ $ as -o bubblesort2.o bubblesort2.s
@ pi@RPi3:~ $ gcc -o bubblesort2 bubblesort2.o
@ pi@RPi3:~ $ ./bubblesort2; echo $?
.data
table:
.word 9, -8, -7, 6, -5, 4, -3, 2, -1, 0, 127, 128
aft_table:
.balign 4
n:
.word (aft_table - table) >>2
@ ---------------
@ ---- Added ----
.balign 4 @ Prt
frmt: @ Prt
.asciz "%d " @ Prt
.balign 4
flags: @ RDmp
.word 0 @ RDmp
msg: @ RDmp
.asciz "NZCV 2> 6= 8<
" @ RDmp
aft_msg: @ RDmp
@ RDmp
l_msg: @ RDmp
.word (aft_msg - msg) - 1 @ RDmp
@ ---------------
.text
.global main
.global printf, putchar @ Prt
main:
push {r4-r10, lr}
ldr r1, =n
ldr r1, [r1] @ len(table) in r1
bl Print @ Prt
@ ---------------
bubblesort:
@ maybe r0, r1 for table and counter (rest is based on that)
@ r8 = Buffer for Value
@ r4 = Counter for loop1
@ r5 = Register for Value
@ r7 = Register for Value
@ r1 = Size of Array
@ r0 = Address of the 1. Element
@ r10 = "swapped" register
loop:
mov r10, #0
sub r1, #1 @ Decrement Counter for loop
@ ---- Added ----
ldr r0, =table
mov r4, #1 @ i = 1 i could = 0, if instructions
@ cmp r4, r1 @ compare r4 and r1
@ add r4, #1 @ Increment counter for loop1
@ were reversed
@ ---------------
loop1:
ldr r5, [r0]
ldr r7, [r0, #4]
cmp r5, r7
movgt r8, r7
movgt r7, r5
movgt r5, r8
strgt r5, [r0]
strgt r7, [r0, #4]
movgt r10, #1
@ ---- Added ----
add r0, #4 @ Index table to next position
@ ---------------
cmp r4, r1 @ compare r4 and r1
add r4, #1 @ Increment counter for loop1
blt loop1 @ is r4 < r1 ? yes = go to loop1
@ ---- Added ----
bl Print @ Prt
bl RDmp @ RDmp
@ ---------------
cmp r10, #1
beq loop
@ ---- Added ----
exit:
pop {r4-r10, lr}
mov r0, #0 @ good return code
bx lr @ Exit with gcc linker
@ ---------------
@ ---- Added ----
Print: @ Prt, table to stdout
push {r0-r10, lr} @ Prt
mrs r10, cpsr @ Prt
ldr r2, =n @ Prt
ldr r2, [r2] @ Prt
ldr r3, =table @ Prt
lp: @ Prt
ldr r0, =frmt @ Prt
ldr r1, [r3], #4 @ Prt
push {r0-r3} @ Prt
bl printf @ Prt
pop {r0-r3} @ Prt
subs r2, #1 @ Prt
bne lp @ Prt
mov r0, #0xa @ Prt
bl putchar @ Prt
msr cpsr_f, r10 @ Prt
pop {r0-r10, lr} @ Prt
mov pc, lr @ Prt
@ ---------------
@ ---- Added ----
RDmp: @ RDmp
push {r0-r10, lr} @ RDmp
mrs r10, cpsr @ RDmp
push {r8-r11} @ RDmp
push {r4-r7} @ RDmp
push {r0-r3} @ RDmp
@ RDmp
ldr r1, =flags @ RDmp
mrs r0, cpsr @ RDmp
str r0, [r1] @ RDmp
b seq @ RDmp
@ RDmp
p_char: @ RDmp
push {r0-r4, lr} @ RDmp
bl putchar @ RDmp
pop {r0-r4, lr} @ RDmp
mov pc, lr @ RDmp
@ RDmp
register: @ RDmp
mov r1, #0 @ RDmp
push {r0, lr} @ RDmp
@ RDmp
nibble: @ RDmp
mov r2, r3 @ RDmp
lsl r2, r1 @ RDmp
lsr r2, #28 @ RDmp
add r2, #48 @ RDmp
cmp r2, #58 @ RDmp
addge r2, #39 @ RDmp
mov r0, r2 @ RDmp
bl p_char @ RDmp
add r1, #4 @ RDmp
cmp r1, #32 @ RDmp
blt nibble @ RDmp
mov r0, #32 @ RDmp
bl p_char @ RDmp
pop {r0, lr} @ RDmp
mov pc, lr @ RDmp
@ RDmp
manage: @ RDmp
push {r0, lr} @ RDmp
mov r3, r4 @ RDmp
bl register @ RDmp
mov r3, r5 @ RDmp
bl register @ RDmp
mov r3, r6 @ RDmp
bl register @ RDmp
mov r3, r7 @ RDmp
bl register @ RDmp
pop {r0, lr} @ RDmp
mov pc, lr @ RDmp
@ RDmp
seq: @ RDmp
pop {r4-r7} @ RDmp
bl manage @ RDmp
mov r0, #32 @ RDmp
bl p_char @ RDmp
pop {r4-r7} @ RDmp
bl manage @ RDmp
mov r0, #0x0a @ RDmp
bl p_char @ RDmp
@ RDmp
pop {r4-r7} @ RDmp
bl manage @ RDmp
mov r0, #32 @ RDmp
bl p_char @ RDmp
ldr r3, =flags @ RDmp
ldr r3, [r3] @ RDmp
bl register @ RDmp
@ RDmp
ldr r2, = l_msg @ RDmp
ldr r2, [r2] @ RDmp
ldr r1, =msg @ RDmp
w_msg: @ RDmp
ldr r0, [r1], #1 @ RDmp
bl p_char @ RDmp
subs r2, #1 @ RDmp
bgt w_msg @ RDmp
msr cpsr_f, r10 @ RDmp
pop {r0-r10, lr} @ RDmp
mov pc, lr @ RDmp
@ ---------------
./bubblesort2 的输出；echo $?，以@"开头的行是我对它们下面的寄存器的标题注释.
The output from ./bubblesort2; echo $?, the lines that start with "@" are my heading comments for the registers below them.
9 -8 -7 6 -5 4 -3 2 -1 0 127 128
-8 -7 6 -5 4 -3 2 -1 0 9 127 128
@ (r0) (r1) (r2) (r3) (r4) (r5) (r6) (r7)
000207f0 0000000b 7ee7077c 00010450 0000000c 0000007f 00010328 00000080
@ (r8) (r9) (r10) (r11) (F)(cpsr) (F) (F val)
00000000 00000000 60000010 00000000 60000010 NZCV 2> 6= 8<
-8 -7 -5 4 -3 2 -1 0 6 9 127 128
000207ec 0000000a 7ee7077c 00010450 0000000b 00000009 00010328 0000007f
00000000 00000000 60000010 00000000 60000010 NZCV 2> 6= 8<
-8 -7 -5 -3 2 -1 0 4 6 9 127 128
000207e8 00000009 7ee7077c 00010450 0000000a 00000006 00010328 00000009
00000000 00000000 60000010 00000000 60000010 NZCV 2> 6= 8<
-8 -7 -5 -3 -1 0 2 4 6 9 127 128
000207e4 00000008 7ee7077c 00010450 00000009 00000004 00010328 00000006
00000000 00000000 60000010 00000000 60000010 NZCV 2> 6= 8<
-8 -7 -5 -3 -1 0 2 4 6 9 127 128
000207e0 00000007 7ee7077c 00010450 00000008 00000002 00010328 00000004
00000000 00000000 60000010 00000000 60000010 NZCV 2> 6= 8<
0
这篇关于ARM 汇编器常规冒泡排序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IT屋！
