Question

我正在将C代码转换为Mips代码

C代码是

int main(void) {
int i;
int data [10] = { 10, -2, 5, 22, 99, 0, -5, 8, 30, 7};
    for(i=0; i<10; i++){
        printf("%d\n", data[i]);
       }
  return 0;
 }

我的Mips代码是

.data
    data: .space 10

    enter: .asciiz "\n"
.text
.globl main
main:
addi $s0, $zero, 10
add $t0, $zero, $zero
sw $s0, data($t0)

addi $s0, $zero, -2
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 5
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 22
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 99
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 0
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, -5
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 8
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 30
addi $t0, $t0, 4
sw $s0, data($t0)

addi $s0, $zero, 7
addi $t0, $t0, 4
sw $s0, data($t0)

addi $t0, $zero, 0
Loop: slti $t1, $t0, 10
      beq $t1, $zero, Exit
      mul $t2, $t0, 4

      li $v0, 1
      lw $a0, data($t2)
      syscall

          li $v0, 4
      la $a0, enter
      syscall

      addi $t0, $t0, 1
      j Loop
Exit:




end:
li $v0, 10
syscall

s1用于保存值。 t0用于索引 t1是slt的标志 t1乘以4是t2

您应该只关注

          li $v0, 4
      la $a0, enter
      syscall

它打印 10-2522990-58307

dddddddddddddddddddddddddddddddddddasdfasdfasdfasdfasdfasdfasdf

请帮助我dddddddddddddddddddddddddddddddddddddd

Answer 1

您没有为data阵列保留足够的内存。 .space 10保留10个字节的内存，但是您正尝试存储10个 words （40个字节）。因此，您最终会覆盖换行符。

如果将声明更改为data: .space 40，则应该获得预期的输出。

我无法在Mips中打印“输入”

1 个答案: