[英]How to reverse an array in assembly language ARM?
(該問題基於匯編語言ARM。)我正在處理一個問題,該問題要求我反轉給定數組。 像這樣:
Given array: 1, 2, 3, 4, 5
Reversed array: 5, 4, 3, 2, 1
這個問題的局限性在於,我只能使用寄存器r0-r3。
我有一個基本的算法,但是當我嘗試實現這個想法時,我真的很困惑。 我的算法:
Loop:
1. get value from head pointer, ptr++
2. get value from tail pointer, ptr--
3. swap them
4. check if head pointer and tail pointer cross,
if so, exit loop and return.
if not, go back to loop.
但我只是不知道如何僅使用4個寄存器來解決此問題。
以下是我目前擁有的所有信息。
.text
.global reverse
reverse:
@ See if head and tail ptr cross
@ If so, end loop (b end)
head:
@ use r2 to represent head value
ldr r2,[r0] @ r2 <-*data get the first value
tail:
@ mov r1,r1 @ size
sub r1,r1,#1 @ size-1
lsl r1,r1,#2 @ (size-1)*4
add r0,r0,r1 @ &data[size-1] need to ldr r1,[r0] to get value
ldr r1,[r0] @ get value for r1 (from tail)
swap:
@ swap values
mov r3, r1 @store value to r3
str r2, [r0]
@ head ptr ++
@ tail ptr --
@ back to reverse
end:
@ loop ends
粗暴和低效的例子
.data
Array: .word 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,32,64,128,256,512
.equ len.Array,.-Array
.text
.global main
main:
nop
sub sp, sp, #len.Array // save space on stack
ldr r1,=Array // Array
mov r2, #len.Array // length of array
mov r3, #0 // zero init counter Array
1:
ldr r0, [r1,r3] // load word size element position x from Array
push {r0} // push element value into stack
add r3, r3, #4 // inc Array counter by 4 since word size is 4 bytes
cmp r3, r2 //
blt 1b
// pop values off the stack - LIFO results in reversal
mov r3, #0 // zero init counter Array
2:
pop {r0} // pop element value from stack - LIFO
str r0, [r1,r3]
add r3, r3, #4 // inc Array counter by 4 since word size is 4 bytes cmp r3, r2
blt 2b add sp, sp, #len.Array // restore stack pointer
GDB輸出:
(gdb) x/21d $r1
0x1102d: 1 2 3 4
0x1103d: 5 6 7 8
0x1104d: 9 10 11 12
0x1105d: 13 14 15 16
0x1106d: 32 64 128 256
0x1107d: 512
(gdb) x/21d $r1
0x1102d: 512 256 128 64
0x1103d: 32 16 15 14
0x1104d: 13 12 11 10
0x1105d: 9 8 7 6
0x1106d: 5 4 3 2
0x1107d: 1
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