加载32位常量以在MIPS中注册 [英] Load 32-bit constant to register in MIPS

问题描述

我在学习MIPS时对这部分感到困惑.

I was confused about this part while I study MIPS.

约翰·轩尼诗(John L. Hennessy)所写的教科书说，如果我们得到一些大常量要加载，我们应该

The textbook written by Professor John L. Hennessy say if we get some big constant to load, we should

lui $s0, upper(big)
ori $s0, $s0, lower(big)

但是我们为什么不做

addi $s0, $zero, big

由于寄存器是32位的，因此更加简单明了，不是吗?

Since the registers are 32-bit, this is more strightforward, isn't it?

推荐答案

传递给addi的直接参数只有16位.要加载超出16位值范围的32位立即数，您需要分两次执行，如教科书中的示例所示.

The immediate argument passed to addi is only 16 bits. To load a 32-bit immediate value that is outside the range of a 16-bit value you need to do it in two goes, as in the example from your text book.

(预料到另一个问题，之所以没有采用32位立即值的立即加载或立即添加指令，是因为MIPS ISA使用固定大小32位指令，因此任何指令参数总是< 32位可用-这在RISC体系结构中非常常见.)

(Anticipating a further question, the reason there is no load immediate or add immediate instruction which takes a 32-bit immediate value is because the MIPS ISA uses fixed size 32-bit instructions, so there are always < 32 bits available for any instruction arguments - this is very common in RISC architectures.)

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