本文介绍了不工作ARM汇编语言菜单的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
/*menu_prompt*/
.balign 4
menu_prompt: .asciz "Choose which problem.\nType 1 for Problem 1\nType 2 for Pr$
/*scanner*/
.balign 4
scan_pattern_int: .asciz "%d"
scan_read: .word 0
.balign 4
return: .word 0
.global main
main:
ldr r0, address_of_return
str lr, [r1]
ldr r0, address_of_menu_prompt
bl printf
ldr r0, address_of_scan_pattern_int
ldr r1, address_of_scan_read
bl scanf
ldr r1, [r1]
str lr, [r1]
ldr r0, address_of_menu_prompt
bl printf
ldr r0, address_of_scan_pattern_int
ldr r1, address_of_scan_read
bl scanf
ldr r1, [r1]
cmp r1, #1
beq in_prob_1
cmp r1, #2
beq in_prob_2
cmp r1, #3
beq in_prob_3
in_prob_1:
mov r0, #100
bal end
in_prob_2:
mov r0, #200
bal end
in_prob_3:
mov r0, #300
bal end
end:
ldr lr, address_of_return
ldr lr, [lr]
bx lr
address_of_menu_prompt: .word menu_prompt
address_of_scan_pattern_int: .word scan_pattern_int
address_of_scan_read: .word scan_read
address_of_return: .word return
后，我没有运行该程序无论我选择哪个输入（1,2或3）$呼应？总是显示R0持有139，我不知道为什么。如果让我选择1 R0应持有100，2 R0应持有200，第3和R0应该持有300
after i run the program no matter which input i choose (1,2, or 3) echo$? always shows that r0 holds 139, i'm not sure why. if i choose 1 r0 should hold 100, 2 r0 should hold 200, 3 and r0 should hold 300.
推荐答案
scanf函数并不需要离开 R1 如你每ARM过程调用标准通过。转移到 scanf函数后，您应该重新加载 R1 。
scanf doesn't need to leave the contents of r1 as you've passed per ARM Procedure Call Standard. You should reload r1 after branching to scanf.
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