Assembly 带post索引偏移量的STR后arm程序停止
我正在完成一个任务，在这个任务中，我需要通过输入接收一个数组，并以相反的方式返回它。我以前问过关于这项作业的问题，这次我快要完成了。我现在的问题是，当我在内存中存储数组的一个值时，我的程序会中途停止。 编辑：我现在粘贴了完整的代码。失败的str与它的崩溃高度相关。对于这个假设，我得到的代码没有主代码，也没有变量primero、segundo和tercero，它们是程序运行时要打印的字符串。打印和FGET是打印到屏幕并从标准DIN读取的功能Assembly 带post索引偏移量的STR后arm程序停止,assembly,arm,undefined-behavior,Assembly,Arm,Undefined Behavior,我正在完成一个任务，在这个任务中，我需要通过输入接收一个数组，并以相反的方式返回它。我以前问过关于这项作业的问题，这次我快要完成了。我现在的问题是，当我在内存中存储数组的一个值时，我的程序会中途停止。 编辑：我现在粘贴了完整的代码。失败的str与它的崩溃高度相关。对于这个假设，我得到的代码没有主代码，也没有变量primero、segundo和tercero，它们是程序运行时要打印的字符串。打印和FGET是打印到屏幕并从标准DIN读取的功能 @ Las siguientes funciones f
@ Las siguientes funciones fueron obtenidas del archivo ejemplo "UsefulFunctions.s" presente
@en la pagina oficial de descarga del software ARMSIM#
@(Agradecimientos al desarrollador original de estas funciones: R.N. Horspool)
.global prints, fprints, fgets
@ PROGRAMA
.text
.global main
main:
ldr r0, =primero
bl prints
ldr r0, adr_num @where n will be stored
mov r1, #4 @buffer for the first number
mov r2, #0 @indicates fgets must read from stdin
bl fgets
bl atoi @turns to int so as to evaluate without caring about a "\n"
mov r5, r0 @store de recived int
ldr r6, =arreglo
firstLoop:
cmp r5, #0
beq msg
ldr r0, =segundo
bl prints
ldr r0, adr_num @where n will be stored
mov r1, #4 @buffer for the first number
mov r2, #0 @indicates fgets must read from stdin
bl fgets
bl atoi @turns to int so as to evaluate without caring about a "\n"
str r0, [r6], #4 @--HERE IT CRASHES--store de recived int in register
sub r5, r5, #1
b firstLoop
msg:
ldr r0, =tercero
bl prints
ldr r0, [r6], #-4
secondLoop:
cmp r8, #0
beq exit
ldr r0, [r6], #-4 @move result to register r0 so as to turn int to string
ldr r1, adr_str @enough space is saved so as to store the result of transformation
bl itoa
bl prints
sub r8, r8, #1
b secondLoop
@Memory data adress in DATOS
adr_num: .word number @Here the space for an int is defined (see end of document, section .DATA)
adr_str: .word string @Here the space for a string is defined (see end of document, section .DATA)
@reading of the first character (n):
ldr r0, adr_num @where n will be stored
mov r1, #4 @buffer for the first number
mov r2, #0 @indicates fgets must read from stdin
bl fgets
bl atoi @turns to int so as to evaluate without caring about a "\n"
mov r5, r0 @store de recived int
mov r0, r5 @move result to register r0 so as to turn int to string
ldr r1, adr_str @enough space is saved so as to store the result of transformation
bl itoa
bl prints
b exit
exit:
mov r0, #0x18
mov r1, #0
swi 0x123456
@ prints: Returns an ASCII string ending in null to stdout
@
@ Abstract use:
@ prints(r0)
@ Inputs:
@ r0: memory adress to ASCII string ending in null
@ Resultado:
@ N/A, but string is writen to stdout (console)
prints:
stmfd sp!, {r0,r1,lr}
ldr r1, =operands
str r0, [r1,#4]
bl strlen
str r0, [r1,#8]
mov r0, #0x0
str r0, [r1]
mov r0, #0x05
swi 0x123456
ldmfd sp!, {r0,r1,pc}
@ fgets: read a line of ASCII text from a stream of inputs (open text file or stdin)
@
@ Abstract use:
@ r0 = fgets(r0, r1, r2)
@ Inputs:
@ r0: memory adress of a buffer where first line will be stored
@ r1: buffer size (must acomodate an ending character)
@ r2: name of a file to open for input or "0" to read from stdin
@ Resultado:
@ r0: buffer memory adress if characters where able to be read or = if @ characters wheren't read by an EOF error.
@ One text line including a terminating linefeed character
@ is read into the buffer, if the buffer is large enough.
@ Otherwise the buffer holds size-1 characters and a null byte.
@ Note: the line stored in the buffer will have only a linefeed
@ (\n) line ending, even if the input source has a DOS line
@ ending (a \r\n pair).
fgets: stmfd sp!, {r1-r4,lr}
ldr r3, =operands
str r2, [r3] @ specify input stream
mov r2, r0
mov r4, r1
mov r0, #1
str r0, [r3,#8] @ to read one character
mov r1, r3
mov r3, r2
1: subs r4, r4, #1
ble 3f @ jump if buffer has been filled
str r3, [r1,#4]
2: mov r0, #0x06 @ read operation
swi 0x123456
cmp r0, #0
bne 4f @ branch if read failed
ldrb r0, [r3]
cmp r0, #'\r' @ ignore \r char (result is a Unix line)
beq 2b
add r3, r3, #1
cmp r0, #'\n'
bne 1b
3: mov r0, #0
strb r0, [r3]
mov r0, r2 @ set success result
ldmfd sp!, {r1-r4,pc}
4: cmp r4, r2
bne 3b @ some chars were read, so return them
mov r0, #0 @ set failure code
strb r0, [r2] @ put empty string in the buffer
ldmfd sp!, {r1-r4,pc}
@ strlen: computes the lenght of a string made form ASCII characters ending in null
@
@ Abstract use:
@ r0 = strlen(r0)
@ Inputs:
@ r0: memory adress of an ASCII string enfing in null.
@ Resultado:
@ r0: string lenght (excluding ending byte)
strlen:
stmfd sp!, {r1-r3,lr}
mov r1, #0
mov r3, r0
1: ldrb r2, [r3], #1
cmp r2, #0
bne 1b
sub r0, r3, r0
ldmfd sp!, {r1-r3,pc}
@ atoi: turns an ASCII string ending in null to it's int equivalent
@
@ Abstract use:
@ r0 = atoi(r0)
@ Inputs:
@ r0: memory adress of an ASCII string ending in null.
@ Resultado:
@ r0: value of te converted int
atoi:
stmfd sp!, {r1-r4,lr}
mov r2, #0 @ holds result
mov r3, #0 @ set to 1 if a negative number
mov r4, #10
1: ldrb r1, [r0], #1 @ get next char
cmp r1, #0
beq 4f
cmp r1, #' '
beq 1b
cmp r1, #'\n' @se añadio la regla para que no procese los '\n'
beq 1b
cmp r1, #'-'
moveq r3, #1
ldreqb r1, [r0], #1
b 3f
2: cmp r1, #9
bgt 4f
mul r2, r4, r2
add r2, r2, r1
ldrb r1, [r0], #1
3: subs r1, r1, #'0'
bge 2b
4: cmp r3, #0
moveq r0, r2
mvnne r0, r2
ldmfd sp!, {r1-r4,pc}
@ itoa: int to ASCII
@
@ Abstract use:
@ r0 = itoa(r0, r1)
@ Exit parameters:
@ r0: signed integer
@ r1: buffer adress that is large enough to keep the functions result, @ @ which will be an ASCII string ending in NULL characterla direccion @ @ de un buffer suficientemente grande para mantener el
@ Resultado:
@ r0: buffers adress
itoa:
stmfd sp!, {r1-r7,lr}
mov r7, r1 @ remember buffer address
cmp r0, #0 @ check if negative and if zero
movlt r2, #'-'
moveq r2, #'0'
strleb r2, [r1],#1 @ store a '-' symbol or a '0' digit
beq 3f
mvnlt r0, r0
ldr r3, =4f @ R3: multiple pointer
mov r6, #0 @ R6: write zero digits? (no, for leading zeros)
1: ldr r4, [r3],#4 @ R4: current power of ten
cmp r4, #1 @ stop when power of ten < 1
blt 3f
mov r5, #0 @ R5: multiples count
2: subs r0, r0, r4 @ subtract multiple from value
addpl r5, r5, #1 @ increment the multiples count
bpl 2b
add r0, r0, r4 @ correct overshoot
cmp r5, #0 @ if digit is '0' and ...
cmpeq r6, #0 @ if it's a leading zero
beq 1b @ then skip it
mov r6, #1
add r2, r5, #'0' @ ASCII code for the digit
strb r2, [r1],#1 @ store it
b 1b
3: mov r0, #0
strb r0, [r1]
mov r0, r7
ldmfd sp!, {r1-r7,pc}
4: .word 1000000000, 100000000, 10000000, 1000000
.word 100000, 10000, 1000, 100, 10, 1, 0
@ DATOS
.data
operands: .word 0, 0, 0
string: .space 32 @buffer for a 32 character string
number: .space 4 @buffer for a number (n y then k) + ending character
list: .space 4000 @buffer for a 1000 numbers max list (only 999 will be used as true max)
found: .ascii "NUmber is in the list."
space: .space 2
unfound: .ascii "Number is not on the list."
primero: .asciz "Enter array lenght: "
segundo: .asciz "Enter next number of the array: "
tercero: .asciz "Inverted array is: "
arreglo: .word 0, 0, 0
多亏了厄尔科林在这一期中的投入，我才能够完成这个项目。最后我用错了STR。在我改变了这一点以及我在数组中移动的方式，并在数组中定义了一个100个字符的空格，如下所示：
.balign 4
array: .skip 400
感谢大家抽出时间只是为了好玩，我将程序转换为使用Linux系统调用：
.syntax unified
.cpu arm1176jzf-s
.arm
@ Las siguientes funciones fueron obtenidas del archivo ejemplo "UsefulFunctions.s" presente
@ en la pagina oficial de descarga del software ARMSIM#
@ (Agradecimientos al desarrollador original de estas funciones: R.N. Horspool)
.global _start
@ PROGRAMA
.text
.type _start, %function
_start:
ldr r0, =primero
bl prints
ldr r0, adr_num @where n will be stored
movs r1, #4 @buffer for the first number
movs r2, #0 @indicates fgets must read from stdin
bl fgets
bl atoi @turns to int so as to evaluate without caring about a "\n"
movs r5, r0 @store de recived int
mov r8, r0
cmp r0, #100
it hi
bhi exit
ldr r6, =arreglo
firstLoop:
cmp r5, #0
beq msg
ldr r0, =segundo
bl prints
ldr r0, adr_num @where n will be stored
movs r1, #4 @buffer for the first number
movs r2, #0 @indicates fgets must read from stdin
bl fgets
bl atoi @turns to int so as to evaluate without caring about a "\n"
str r0, [r6], #4 @--HERE IT CRASHES--store de recived int in register
sub r5, r5, #1
b firstLoop
msg:
ldr r0, =tercero
bl prints
secondLoop:
cmp r8, #0
beq exit
ldr r0, [r6, #-4]! @move result to register r0 so as to turn int to string
ldr r1, adr_str @enough space is saved so as to store the result of transformation
bl itoa
bl prints
sub r8, r8, #1
b secondLoop
@Memory data adress in DATOS
adr_num: .word number @Here the space for an int is defined (see end of document, section .DATA)
adr_str: .word string @Here the space for a string is defined (see end of document, section .DATA)
.type exit, %function
exit:
movs r7, #1
movs r0, #0
svc #0
@ prints: Returns an ASCII string ending in null to stdout
@
@ Abstract use:
@ prints(r0)
@ Inputs:
@ r0: memory adress to ASCII string ending in null
@ Resultado:
@ N/A, but string is writen to stdout (console)
.type prints, %function
prints:
stmfd sp!, {r0-r7,lr}
bl strlen
movs r7, #4
movs r2, r0
ldr r1, [sp]
movs r0, #1
svc #0
ldmfd sp!, {r0-r7,pc}
@ fgets: read a line of ASCII text from a stream of inputs (open text file or stdin)
@
@ Abstract use:
@ r0 = fgets(r0, r1, r2)
@ Inputs:
@ r0: memory adress of a buffer where first line will be stored
@ r1: buffer size (must acomodate an ending character)
@ r2: name of a file to open for input or "0" to read from stdin
@ Resultado:
@ r0: buffer memory adress if characters where able to be read or = if @ characters wheren't read by an EOF error.
@ One text line including a terminating linefeed character
@ is read into the buffer, if the buffer is large enough.
@ Otherwise the buffer holds size-1 characters and a null byte.
@ Note: the line stored in the buffer will have only a linefeed
@ (\n) line ending, even if the input source has a DOS line
@ ending (a \r\n pair).
.type fgets, %function
fgets: stmfd sp!, {r0-r9,lr}
mov r8, r1 @ Count
mov r9, r0 @ Buffer
1: subs r8, r8, #1
ble 3f @ jump if buffer has been filled
2:
ldr r0, [sp, #8] @ File Descriptor
mov r1, r9 @ Buffer
movs r2, #1 @ Count
movs r7, #3 @ read syscall
svc #0
cmp r0, #1
bne 4f @ branch if read failed
ldrb r0, [r9]
cmp r0, #'\r' @ ignore \r char (result is a Unix line)
beq 2b
adds r9, r9, #1
cmp r0, #'\n'
bne 1b
3: movs r0, #0
strb r0, [r9]
ldmfd sp!, {r0-r9,pc}
4: ldr r2, [sp, #4]
cmp r8, r2
bne 3b @ some chars were read, so return them
movs r0, #0 @ set failure code
strb r0, [r9] @ put empty string in the buffer
adds sp, #4 @ Skip over saved buffer pointer
ldmfd sp!, {r1-r9,pc}
@ strlen: computes the lenght of a string made form ASCII characters ending in null
@
@ Abstract use:
@ r0 = strlen(r0)
@ Inputs:
@ r0: memory adress of an ASCII string enfing in null.
@ Resultado:
@ r0: string lenght (excluding ending byte)
.type strlen, %function
strlen:
stmfd sp!, {r1-r2,lr}
mov r2, r0
1: ldrb r1, [r2], #1
tst r1, r1
bne 1b
sub r0, r2, r0
ldmfd sp!, {r1-r2,pc}
@ atoi: turns an ASCII string ending in null to it's int equivalent
@
@ Abstract use:
@ r0 = atoi(r0)
@ Inputs:
@ r0: memory adress of an ASCII string ending in null.
@ Resultado:
@ r0: value of te converted int
.type atoi, %function
atoi:
stmfd sp!, {r1-r4,lr}
movs r2, #0 @ holds result
movs r3, #0 @ set to 1 if a negative number
movs r4, #10
1: ldrb r1, [r0], #1 @ get next char
cmp r1, #0
beq 4f
cmp r1, #' '
beq 1b
cmp r1, #'\n' @se añadio la regla para que no procese los '\n'
beq 1b
cmp r1, #'-'
itt eq
moveq r3, #1
ldrbeq r1, [r0], #1
b 3f
2: cmp r1, #9
bgt 4f
mul r2, r4, r2
add r2, r2, r1
ldrb r1, [r0], #1
3: subs r1, r1, #'0'
bge 2b
4: cmp r3, #0
ite eq
moveq r0, r2
rsbne r0, r2, #0
ldmfd sp!, {r1-r4,pc}
@ itoa: int to ASCII
@
@ Abstract use:
@ r0 = itoa(r0, r1)
@ Exit parameters:
@ r0: signed integer
@ r1: buffer adress that is large enough to keep the functions result, @ @ which will be an ASCII string ending in NULL characterla direccion @ @ de un buffer suficientemente grande para mantener el
@ Resultado:
@ r0: buffers adress
itoa:
stmfd sp!, {r1-r7,lr}
mov r7, r1 @ remember buffer address
cmp r0, #0 @ check if negative and if zero
it lt
movlt r2, #'-'
it eq
moveq r2, #'0'
it le
strble r2, [r1],#1 @ store a '-' symbol or a '0' digit
beq 3f
it lt
rsblt r0, r0, #0
ldr r3, =4f @ R3: multiple pointer
movs r6, #0 @ R6: write zero digits? (no, for leading zeros)
1: ldr r4, [r3],#4 @ R4: current power of ten
cmp r4, #1 @ stop when power of ten < 1
blt 3f
movs r5, #0 @ R5: multiples count
2: subs r0, r0, r4 @ subtract multiple from value
itt pl
addpl r5, r5, #1 @ increment the multiples count
bpl 2b
add r0, r0, r4 @ correct overshoot
cmp r5, #0 @ if digit is '0' and ...
itt eq
cmpeq r6, #0 @ if it's a leading zero
beq 1b @ then skip it
movs r6, #1
add r2, r5, #'0' @ ASCII code for the digit
strb r2, [r1],#1 @ store it
b 1b
3: movs r0, #0
strb r0, [r1]
movs r0, r7
ldmfd sp!, {r1-r7,pc}
4: .word 1000000000, 100000000, 10000000, 1000000
.word 100000, 10000, 1000, 100, 10, 1, 0
@ DATOS
.data
.align 4
.type string, %object
string: .space 32 @buffer for a 32 character string
.type number, %object
number: .space 4 @buffer for a number (n y then k) + ending character
.type list, %object
list: .space 4000 @buffer for a 1000 numbers max list (only 999 will be used as true max)
.type found, %object
found: .ascii "NUmber is in the list."
.align 4
.type unfound, %object
unfound: .ascii "Number is not on the list."
.align 4
.type primero, %object
primero: .asciz "Enter array lenght: "
.align 4
.type segundo, %object
segundo: .asciz "Enter next number of the array: "
.align 4
.type tercero, %object
tercero: .asciz "Inverted array is: "
.align 4
.type arreglo, %object
arreglo: .space 100*4
armlinux-gnueabihf-as-g test.S-o test.o&&armlinux-gnueabihf-ld test.o-o test&&qemu-arm./test
qemu arm-g 1234./test运行，您可以通过GDB调试它：
$gdb多拱。/test
（gdb）目标扩展远程：1234
使用：1234进行远程调试
_从测试开始（）。S:16
16 ldr r0，=primero
（gdb）stepi
...
在基于ARM的SBC（如Raspberry PI）上，您可以通过安装binutils
对其进行测试，然后：
as -g test.S -o test.o && ld test.o -o test && ./test
什么是打印
，primero
，adr_num
，arreglo
。。。？请出示整个节目。任何地方都没有push{lr}
，因此您不能从main
返回，这是故意的吗？由于您显然正在使用C库，因此只需从main
返回（bx-lr
），而无需调用exit
。此外，对于fgets
，您不能在r2
中通过0
；您必须加载stdin
变量的地址（可能是ldr r2，=stdin
）。lr
的处理应该在她/他调用beq exit
时起作用。给出数据声明和代码会很有帮助。如果我们被告知哪个str
失败了，这将非常有帮助。寄存器是无效内存，您得到了SIGSEGV或类似的东西itoa（）
通常需要三个参数。如果arreglo
在.text段中，它将在str
上崩溃。请详细说明问题，并说明到目前为止为调试所做的工作this@old_timer我添加了完整的程序并解释了一些事情。感谢您的输入。您介意将注释和变量名翻译成英语吗？ARM具有缩放索引寻址模式（左移）；如果你是这个意思的话，你不需要单独乘法。
as -g test.S -o test.o && ld test.o -o test && ./test