树莓派4 32位ARM汇编程序运算结果全为0问题求助

核心问题分析 #

你的代码存在两个关键错误，导致运算结果无法正确输出：

1. printf参数传递不完整：运算结果的格式化字符串（如sum_message）包含3个%d占位符，需要传入3个整数参数（输入值1、输入值2、运算结果），但你调用printf时仅传入了前两个参数，第三个参数未设置，导致读取到寄存器的默认残留值（此处为0）。
2. 非易失寄存器未遵循调用规范：ARM AAPCS调用标准中，r4-r6属于非易失寄存器，函数使用这些寄存器前必须将其压栈保存，避免被其他函数（如printf）破坏。

修正后的代码 #

/* -- two_int_data_op.s -- */

.section .data
/* Prompt message */
prompt: .asciz "Please enter two signed integers: "

/* Response message */
response: .asciz "You entered %d and %d from the keyboard, now some operations on those values!\n"
sum_message: .asciz "Sum of %d and %d is %d\n" /*print out the sum of chosen values*/
product_message: .asciz "Product of %d and %d is %d\n" /*print out the product of chosen values*/
logical_and_message: .asciz "Logical AND of %d and %d is %d\n" /*print out the logical AND of chosen values*/
logical_or_message: .asciz "Logical OR of %d and %d is %d\n" /*print out the logical OR chosen values*/

/* Format pattern for scanf */
pattern: .asciz "%d %d"

/* Where scanf will store the number read */
value_read1: .word 0
value_read2: .word 0

.section .text
.global main
main:
        push {r4, r5, r6, lr}  /* 保存非易失寄存器和返回地址 */

        // use r4, r5 as registers holding pointers to value_read1, and value_read2
        ldr r4, =value_read1
        ldr r5, =value_read2

        ldr r0, =prompt         /* r0 contains pointer to prompt message */
        bl printf               /* call printf to output our prompt */

        ldr r0, =pattern        /* r0 contains pointer to format string for our scan pattern */
        mov r1, r4              /* r1 contains pointer to variable label where our first number is stored */
        mov r2, r5              /* r2 contains pointer to variable label where our second number is stored */
        bl scanf                /* call to scanf */

next:
        ldr r0, =response       /* r0 contains pointer to response message */
        mov r1, r4              /* r4 contains pointer to value_read1 */
        ldr r1, [r1]            /* r1 contains value dereferenced from r1 in previous instruction */
        mov r2, r5
        ldr r2, [r2]
        bl printf               /* call printf to output our response */

        // 加法运算与输出
        mov r1, r4              /* r4 contains pointer to value_read1 */
        ldr r1, [r1]            /* r1 contains first input value */
        mov r2, r5
        ldr r2, [r2]            /* r2 contains second input value */
        add r6, r1, r2          /* calculate the sum of the two integers, store result r6 */
        ldr r0, =sum_message    /* r0 contains pointer to sum message */
        mov r3, r6              /* 将运算结果传入第三个参数寄存器r3 */
        bl printf               /* call printf to output sum */

        // 乘法运算与输出
        mov r1, r4              /* r4 contains pointer to value_read1*/
        ldr r1, [r1]            /* r1 contains first input value */
        mov r2, r5
        ldr r2, [r2]            /* r2 contains second input value */
        mul r6, r1, r2          /* calculate the product of the two integers, store result r6 */
        ldr r0, =product_message /* r0 contains pointer to product message*/
        mov r3, r6              /* 将运算结果传入第三个参数寄存器r3 */
        bl printf               /* call printf to output product */

        // 按位与运算与输出
        mov r1, r4              /* r4 contains pointer to value_read1*/
        ldr r1, [r1]            /* r1 contains first input value */
        mov r2, r5
        ldr r2, [r2]            /* r2 contains second input value */
        and r6, r1, r2          /* bitwise and r1 and r2, store result in r6*/
        ldr r0, =logical_and_message    /* r0 contains pointer to logical and message*/
        mov r3, r6              /* 将运算结果传入第三个参数寄存器r3 */
        bl printf

        // 按位或运算与输出
        mov r1, r4              /* r4 contains pointer value_read1 */
        ldr r1, [r1]            /* r1 contains first input value */
        mov r2, r5
        ldr r2, [r2]            /* r2 contains second input value */
        orr r6, r1, r2          /* bitwise or r1, r2, store result in r6 */
        ldr r0, =logical_or_message     /* r0 contains pointer to logical or message */
        mov r3, r6              /* 将运算结果传入第三个参数寄存器r3 */
        bl printf

        mov r0, #0              /* exit code 0 = program terminated normally */
        pop {r4, r5, r6, pc}    /* 恢复寄存器并退出 */

关键修改说明 #

• 寄存器保存：在main函数开头将r4、r5、r6与lr一起压栈，结尾恢复，符合ARM调用规范。
• 参数补全：每次调用printf输出运算结果前，将存放在r6的运算结果赋值给r3（printf的第三个参数寄存器），确保格式化字符串的所有占位符都能获取到正确的值。
• （额外优化）移除了格式化字符串末尾多余的空格，避免输出不必要的空白。

内容的提问来源于stack exchange，提问作者Vincent_Matthew