ARM 汇编语言菜单不起作用

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

在我运行程序后，无论我选择哪个输入(1,2 或 3)echo$?总是显示 r0 为 139，我不知道为什么.如果我选择 1 r0 应该持有 100，2 r0 应该持有 200，3 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 不需要保留 r1 的内容，因为你已经通过了每个 ARM 程序呼叫标准.您应该在分支到 scanf 后重新加载 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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