如何制作裸机 ARM 程序并在 QEMU 上运行它们? [英] How to make bare metal ARM programs and run them on QEMU?
问题描述
我试图让这个 tutorial 按预期工作成功(在 bl main 指令之后失败).根据教程命令
I am trying to get this tutorial to work as intended without success (Something fails after the bl main instruction). According to the tutorial the command
(qemu) xp /1dw 0xa0000018
应该导致打印 33(但我得到 0x00 代替)
should result in the print 33 (But i get 0x00 instead)
a0000018: 33
这是主调用后寄存器的内容(见startup.s)
This is the content of the registers after the main call (see startup.s)
(qemu) info registers
R00=a000001c R01=a000001c R02=00000006 R03=00000000
R04=00000000 R05=00000005 R06=00000006 R07=00000007
R08=00000008 R09=00000009 R10=00000000 R11=a3fffffc
R12=00000000 R13=00000000 R14=0000003c R15=00000004
PSR=800001db N--- A und32
FPSCR: 00000000
我有以下文件
main.c
startup.s
lscript.ld
Makefile
我正在使用以下工具链
arm-2013.11-24-arm-none-eabi-i686-pc-linux-gnu
生成文件:
SRCS := main.c startup.s
LINKER_NAME := lscript.ld
ELF_NAME := program.elf
BIN_NAME := program.bin
FLASH_NAME := flash.bin
CC := arm-none-eabi
CFLAGS := -nostdlib
OBJFLAGS ?= -DS
QEMUFLAGS := -M connex -pflash $(FLASH_NAME) -nographic -serial /dev/null
# Allocate 16MB to use as a virtual flash for th qemu
# bs = blocksize -> 4KB
# count = number of block -> 4096
# totalsize = 16MB
setup:
dd if=/dev/zero of=$(FLASH_NAME) bs=4096 count=4096
# Compile srcs and write to virtual flash
all: clean setup
$(CC)-gcc $(CFLAGS) -o $(ELF_NAME) -T $(LINKER_NAME) $(SRCS)
$(CC)-objcopy -O binary $(ELF_NAME) $(BIN_NAME)
dd if=$(BIN_NAME) of=$(FLASH_NAME) bs=4096 conv=notrunc
objdump:
$(CC)-objdump $(OBJFLAGS) $(ELF_NAME)
mem-placement:
$(CC)-nm -n $(ELF_NAME)
qemu:
qemu-system-arm $(QEMUFLAGS)
clean:
rm -rf *.bin
rm -rf *.elf
main.c:
static int arr[] = { 1, 10, 4, 5, 6, 7 };
static int sum;
static const int n = sizeof(arr) / sizeof(arr[0]);
int main()
{
int i;
for (i = 0; i < n; i++){
sum += arr[i];
}
return 0;
}
startup.s:
.section "vectors"
reset: b _start
undef: b undef
swi: b swi
pabt: b pabt
dabt: b dabt
nop
irq: b irq
fiq: b fiq
.text
_start:
init:
@@ Copy data to RAM.
ldr r0, =flash_sdata
ldr r1, =ram_sdata
ldr r2, =data_size
@@ Handle data_size == 0
cmp r2, #0
beq init_bss
copy:
ldrb r4, [r0], #1
strb r4, [r1], #1
subs r2, r2, #1
bne copy
init_bss:
@@ Initialize .bss
ldr r0, =sbss
ldr r1, =ebss
ldr r2, =bss_size
@@ Handle bss_size == 0
cmp r2, #0
beq init_stack
mov r4, #0
zero:
strb r4, [r0], #1
subs r2, r2, #1
bne zero
init_stack:
@@ Initialize the stack pointer
ldr sp, =0xA4000000
@@ **this call dosent work as expected.. (r13/sp contains 0xA4000000)**
bl main
@@ Dosent return from main
@@ r0 should now contain 33
stop:
b stop
lscript.ld:
/*
* Linker for testing purposes
* (using 16 MB virtual flash = 0x0100_0000)
*/
MEMORY {
rom (rx) : ORIGIN = 0x00000000, LENGTH = 0x01000000
ram (rwx) : ORIGIN = 0xA0000000, LENGTH = 0x04000000
}
SECTIONS {
.text : {
* (vectors);
* (.text);
} > rom
.rodata : {
* (.rodata);
} > rom
flash_sdata = .;
ram_sdata = ORIGIN(ram);
.data : AT (flash_sdata) {
* (.data);
} > ram
ram_edata = .;
data_size = ram_edata - ram_sdata;
sbss = .;
.bss : {
* (.bss);
} > ram
ebss = .;
bss_size = ebss - sbss;
/DISCARD/ : {
*(.note*)
*(.comment)
*(.ARM*)
/*
*(.debug*)
*/
}
}
可执行文件的反汇编(objdump):
program.elf: file format elf32-littlearm
Disassembly of section .text:
00000000 <reset>:
0: ea000023 b 94 <_start>
00000004 <undef>:
4: eafffffe b 4 <undef>
00000008 <swi>:
8: eafffffe b 8 <swi>
0000000c <pabt>:
c: eafffffe b c <pabt>
00000010 <dabt>:
10: eafffffe b 10 <dabt>
14: e320f000 nop {0}
00000018 <irq>:
18: eafffffe b 18 <irq>
0000001c <fiq>:
1c: eafffffe b 1c <fiq>
00000020 <main>:
20: e52db004 push {fp} ; (str fp, [sp, #-4]!)
24: e28db000 add fp, sp, #0
28: e24dd00c sub sp, sp, #12
2c: e3a03000 mov r3, #0
30: e50b3008 str r3, [fp, #-8]
34: ea00000d b 70 <main+0x50>
38: e3003000 movw r3, #0
3c: e34a3000 movt r3, #40960 ; 0xa000
40: e51b2008 ldr r2, [fp, #-8]
44: e7932102 ldr r2, [r3, r2, lsl #2]
48: e3003018 movw r3, #24
4c: e34a3000 movt r3, #40960 ; 0xa000
50: e5933000 ldr r3, [r3]
54: e0822003 add r2, r2, r3
58: e3003018 movw r3, #24
5c: e34a3000 movt r3, #40960 ; 0xa000
60: e5832000 str r2, [r3]
64: e51b3008 ldr r3, [fp, #-8]
68: e2833001 add r3, r3, #1
6c: e50b3008 str r3, [fp, #-8]
70: e3a02006 mov r2, #6
74: e51b3008 ldr r3, [fp, #-8]
78: e1530002 cmp r3, r2
7c: baffffed blt 38 <main+0x18>
80: e3a03000 mov r3, #0
84: e1a00003 mov r0, r3
88: e24bd000 sub sp, fp, #0
8c: e49db004 pop {fp} ; (ldr fp, [sp], #4)
90: e12fff1e bx lr
00000094 <_start>:
94: e59f004c ldr r0, [pc, #76] ; e8 <stop+0x4>
98: e59f104c ldr r1, [pc, #76] ; ec <stop+0x8>
9c: e59f204c ldr r2, [pc, #76] ; f0 <stop+0xc>
a0: e3520000 cmp r2, #0
a4: 0a000003 beq b8 <init_bss>
000000a8 <copy>:
a8: e4d04001 ldrb r4, [r0], #1
ac: e4c14001 strb r4, [r1], #1
b0: e2522001 subs r2, r2, #1
b4: 1afffffb bne a8 <copy>
000000b8 <init_bss>:
b8: e59f0034 ldr r0, [pc, #52] ; f4 <stop+0x10>
bc: e59f1034 ldr r1, [pc, #52] ; f8 <stop+0x14>
c0: e59f2034 ldr r2, [pc, #52] ; fc <stop+0x18>
c4: e3520000 cmp r2, #0
c8: 0a000003 beq dc <init_stack>
cc: e3a04000 mov r4, #0
000000d0 <zero>:
d0: e4c04001 strb r4, [r0], #1
d4: e2522001 subs r2, r2, #1
d8: 1afffffc bne d0 <zero>
000000dc <init_stack>:
dc: e3a0d329 mov sp, #-1543503872 ; 0xa4000000
e0: ebffffce bl 20 <main>
000000e4 <stop>:
e4: eafffffe b e4 <stop>
e8: 00000104 andeq r0, r0, r4, lsl #2
ec: a0000000 andge r0, r0, r0
f0: 00000018 andeq r0, r0, r8, lsl r0
f4: a0000018 andge r0, r0, r8, lsl r0
f8: a000001c andge r0, r0, ip, lsl r0
fc: 00000004 andeq r0, r0, r4
Disassembly of section .rodata:
00000100 <n>:
100: 00000006 andeq r0, r0, r6
Disassembly of section .data:
a0000000 <arr>:
a0000000: 00000001 andeq r0, r0, r1
a0000004: 0000000a andeq r0, r0, sl
a0000008: 00000004 andeq r0, r0, r4
a000000c: 00000005 andeq r0, r0, r5
a0000010: 00000006 andeq r0, r0, r6
a0000014: 00000007 andeq r0, r0, r7
Disassembly of section .bss:
a0000018 <sum>:
a0000018: 00000000 andeq r0, r0, r0
有人能指出正确的方向吗?为什么这不符合我的期望?
Can someone point me in the right direction to why this isn't working according to my expectations?
谢谢亨里克
推荐答案
调试中！
首先，查看 PC 和 PSR:您处于 Undef 模式，处于未定义的指令处理程序中.
First, look at the PC and PSR: You're in Undef mode, in the undefined instruction handler.
好的，在异常模式下，LR 会告诉您异常发生的位置.PC 偏移量和首选返回地址之间有一些稍微复杂的规则，确切地 确定它指向的位置，但只要看一眼，它显然就在 movw/movt 对.
OK, in an exception mode, the LR tells you where you took the exception. There are some slightly complicated rules between the PC offset and the preferred return address determining exactly what it points at, but just eyeballing it it's clearly in the vicinity of the movw/movt pair.
movw 指令实际上只存在于 ARMv7 ISA 及更高版本中.一个简短的调查告诉我你正在模拟的机器是一些旧的 PXA255 东西，它的 CPU 只实现了 ARMv5 ISA.因此，它在早于多年的指令上出现错误也就不足为奇了.
The movw instruction effectively only exists in the ARMv7 ISA onwards. A brief investigation tells me the machine you're emulating is some old PXA255 thing, whose CPU only implements the ARMv5 ISA. Thus it's not surprising it faults on an instruction that it predates by many years.
您的编译器显然已默认配置为以 ARMv7 为目标(这并不罕见)，因此您需要至少将 -march=armv5te 添加到您的 CFLAGS 以针对适当的体系结构版本.高级"挑战将是切换到不同的、更新的机器，但这将涉及使链接描述文件适应新的内存映射并为新的外围设备重写任何涉及硬件的代码，所以我会把这个想法留到从长远来看，一旦您对裸机代码的基础知识感到满意并仔细阅读硬件参考手册.
Your compiler is apparently configured to target ARMv7 by default (which is not uncommon), so you need to add at least -march=armv5te to your CFLAGS to target the appropriate architecture version. The 'advanced' challenge would be to switch to a different, newer, machine, but that's going to involve adapting the linker script to a new memory map and rewriting any hardware-touching code for new peripherals, so I'd save that idea for the longer term, once you're comfortable with the basics of bare-metal code and slogging through hardware reference manuals.
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