ARM汇编：如何输出存储在寄存器中的值

问：

我正在为 raspberry pi 4 32 位编写 ARM 代码。我需要让用户输入两个有符号的 int 值，然后使用这些值来计算运算 add、mul 和 orr 并输出到屏幕。当我运行程序时，每次操作的结果都是 0。分配是显示所选的用户输入以及这些输入的操作结果。

附件是程序的输出。

在此处输入图像描述

这是我的代码：

/* -- 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 {lr}               /* save our return address */

        // 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 */

        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]
        add r6, r1, r2          /* calculate the sum of the two integers, store result r6 */
        ldr r0, =sum_message    /* r0 contains pointer to sum  message */
        bl printf               /* call printf to output our response */

        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]
        mul r6, r1, r2          /* calculate the product of  the two integers, store result r6 */
        ldr r0, =product_message /* r0 contains pointer to product message*/
        bl printf                       /* call printf to output our response */

        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]
        and r6, r1, r2                  /* bitwise and r1 and r2, store result in r6*/
        ldr r0, =logical_and_message    /* r0 contains pointer to logical and mesage*/
        bl printf

        mov r1, r4                      /* r4 cntains pointer value_read1 */
        ldr r1, [r1]                    /* r1 contains value dereferenced from r1 in previous instruction */
        mov r2, r5
        ldr r2, [r2]
        orr r6, r1, r2                  /* bitwise or r1, r2, sotre result in r6 */
        ldr r0, =logical_or_message     /* ro contains pointer to logical or message */
        bl printf

        mov r0, #0              /* exit code 0 = program terminated normally */
        pop {pc}                /* exit our main function */

我期望 r6 输出操作的计算值。相反，所有 4 个操作都会导致 0。