Question

我试图提示用户输入两个整数，然后打印这两个整数的总和。我已经弄清楚如何打印一个整数n，但我无法打印第二个整数，m。

当我运行此代码时，它按预期工作，除了它只打印n的值。

以下是步骤和我的代码：

/* 1. Prompt the user to enter an integer
* 2. Read an integer from the keyboard into memory
* 3. Prompt the user to enter another integer
* 4. Read an integer from the keyboard into memory
* 5. Load the two integers into CPU registers
* 6. Add them together
* 7. Print the result
*/

            .data
    prompt: .asciz  "Enter a number: "  @ user prompt
            .align  2
    sformat:.asciz  "%d"                @ Format string for reading an int with scanf
            .align  2
    pformat:.asciz  "The sum is: %d\n"  @ Format string for printf
            .align  2
    n:      .word   0                   @ int n = 0
    m:      .word   0                   @ int m = 0

            .text
            .global main
    main:   stmfd   sp!, {lr}       @ push lr onto stack

            @ printf("Enter a number: ")
            ldr     r0, =prompt
            bl      printf

            @ scanf("%d\0", &n)
            ldr     r0, =sformat    @ load address of format string
            ldr     r1, =n          @ load address of int n variable
            bl      scanf           @ call scanf("%d", &n)

            @ printf("Enter a number: ")
            ldr     r0, =prompt
            bl      printf

            @ scanf("%d\0" &m)
            ldr     r0, =sformat    @ load address of format string
            ldr     r2, =m          @ load address of int m variable
            bl      scanf           @ call scanf("%d", &m)

            @ printf("You entered %d\n", n)
            ldr     r0, =pformat    @ load address of format string
            ldr     r2, =m
            ldr     r2, [r2]
            ldr     r1, =n          @ load address of int variable
            ldr     r1, [r1]        @ load int variable
            add     r1, r2, r1
            bl      printf          @ call printf("You entered %d\n", n)

            ldmfd   sp!, {lr}       @ pop lr from stack
            mov     r0, #0          @ load return value
            mov     pc, lr          @ return from main
            .end

编辑：我改变了这段代码，使用r1而不是r2：

@ scanf("%d\0" &m)
ldr     r0, =sformat    @ load address of format string
ldr     r2, =m          @ load address of int m variable
bl      scanf           @ call scanf("%d", &m)

这有效，但我不明白为什么。

Answer 1

您的修复工作正常，因为所有体系结构都遵循某种调用约定，其中寄存器映射到函数参数并返回值。在这种情况下，scanf接受两个参数，因此根据ARM EABI calling convention ，第一个参数将在r0中，第二个参数在r1中。

在ARM程序集中添加和打印两个数字

1 个答案: