.space在mips中有什么作用? [英] what does .space do in mips?

问题描述

我遇到了一个分配问题，在该分配中，我们将这些数字放入了一个数组中，并在不使用循环的情况下将它们相加.我已经解决了这个问题，但是.space使我感到困惑.通过设置.space 20，我可以为5个单词腾出空间还是在做其他事情.

I got this problem for an assignment in which we have put these number in an array and add them without using a loop. I have solved this problem but .space is confusing me. By making .space 20 do i make space for 5 words or is it doing something else.

 .data
  array: .space 20
 .text
addi $s0, $zero, 2
addi $s1, $zero, 12
addi $s2, $zero, -5
addi $s3, $zero, 7
addi $s4, $zero, 4
  addi $t0, $zero,0  #index initialized at 0
sw $s0,array($t0) 
addi $t0, $t0, 4
sw $s1,array($t0) 
addi $t0, $t0, 4
sw $s2,array($t0) 
addi $t0, $t0, 4
sw $s3,array($t0) 
addi $t0, $t0, 4
sw $s4,array($t0) 
addi $t0, $t0, 4

推荐答案

.space Len指令指示汇编程序保留Len字节. 由于每个字都有4个字节，因此当Len为20时，您将指示汇编程序保留5个字.

.space Len directive instructs the assembler to reserve Len bytes. As every word has 4 bytes, when Len is 20 you are instructing the assembler to reserve 5 words.

例如，如果您有

.data
array: .space 20
other_data: .asciiz 'This is other data'

然后other_data将是array地址之后的20个字节.

then other_data will be 20 bytes after array address.

由于MIPS中的体系结构限制，如果您想逐字访问(在您的特定示例中，您可能还需要指示汇编器在数组标签之前对齐保留的内存(.align 2)).不需要它，它应该已经对齐了.

Due to architectural constraints in MIPS you may need to also instruct the assembler to align the reserved memory (.align 2) before your array label if you want to access on a word-by-word basis (in your particular example you would not need it, it should be already aligned).

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