MIPS组件，Lui 0x1001 [英] MIPS Assembly, lui 0x1001
问题描述
我有一个作业，其中我必须解释有关以下MIPS汇编代码的一些事情:
I have an assignment in which I have to explain some things about the following MIPS Assembly code:
.data
x: .word 4711
y: .word 10
z: .word 0x0A91
e: .word 0
.text
.globl main
main:
lw $2, x
lw $3, y
lw $4, z
add $2, $2, $3
sub $3, $2, $4
sw $3, e
li $2, 10
syscall
在汇编时，第一条指令lw $2, x被分为两条指令.指令是lui $1, 0x00001001，后跟lw $2, 0x00000000($1).我知道lui将十六进制值1001移到寄存器的上部，此时存储在$ 1中的值为0x10010000，但我不知道1001的来源以及第二条指令的含义.我真的很乐意为您提供帮助.我正在使用MARS来组装和运行此程序.
The first instruction lw $2, x is separated into two instructions when assembled. The instructions are lui $1, 0x00001001 followed by lw $2, 0x00000000($1). I understand that lui moves the hex value 1001 into the upper part of the register and the value stored in $1 at this point is 0x10010000, but I do not understand where the 1001 comes from and what the second instruction means at all. I would really appreacitate any help on the subject.I am using MARS to assemble and run this program.
推荐答案
MIPS指令的长度为32位，程序使用的地址也是如此.
这意味着 lw 指令无法执行将完整的32位地址指定为立即数.
简而言之，指令lw $t, var无效(仅在少数情况下有效).
MIPS instructions are 32 bits longs, and so are the addresses uses by a program.
This implies that the lw instruction is unable to specify a full 32-bit address as an immediate.
Simply put, the instruction lw $t, var is not valid (expect for very few cases).
实际上，它的编码是
lw $t, offset($s)
1000 11ss ssst tttt iiii iiii iiii iiii
i 位显示只有16位用于指定地址(并且必须始终指定基址寄存器，最终可以使用$zero寄存器).
Where the i bits show that only 16 bits are used to specify an address (and that a base register must always be specified, eventually the $zero register can be used).
因此，汇编程序会执行此技巧:每当您使用lw $t, var时，它将把该指令汇编为两条指令，其中一条将地址的高16位装入$at和$at作为基址寄存器，地址的低16位作为偏移量.
So the assembler does this trick: whenever you use a lw $t, var it assembles that instruction into two instructions, one that load the upper 16 bits of the address into $at and a lw that use $at as a base register with the lower 16 bits of the address as the offset.
lui $at, ADDR_H #ADDR_H is ADDR >> 16
lw $t, ADDR_L($at) #ADDR_L is ADDR & 0xffff
请注意，由于lw从$at + ADDR_L 中读取，因此最终使用的地址为 ADDR_H << 16 + ADDR_L = ADDR .如预期的那样.
Note that since the lw reads from $at + ADDR_L the final address used is ADDR_H << 16 + ADDR_L = ADDR. As expected.
这里有个微妙之处，由迈克·斯皮维(非常感谢他)，请参见下文
There is subtlety here, pointed out by Mike Spivey (Many thanks to him), see below
这种不会直接映射到ISA中的指令称为 pseudo-instruction .
$at寄存器保留给汇编器以实现它们.
This kind of instructions, that doesn't map directly into the ISA, are called pseudo-instruction.
The $at register is reserved for the assembler exactly for implementing them.
在MARS中，您可以通过取消选中设置>允许扩展(伪)指令和格式来禁用伪指令.
尽管没有伪指令的编程将很快变得令人讨厌，但要完全了解MIPS架构，至少值得做一次.
In MARS you can disable the pseudo instructions by unchecking Settings > Permits extended (pseudo) instructions and format.
While programming without pseudo-instructions will grow annoying pretty quickly, it is worth doing at least once, to fully understand the MIPS architecture.
Mike Spivey 正确地指出了16位立即偏移量是 sign-extended ，然后再添加到基本寄存器中.
如果ADDR_L被解释为16位二进制补码，结果是负数，则需要更正称为ADDR_H的值.
如果事实如此，则必须将ADDR_H递增.
可以将ADDR_H的通用公式校正为ADDR_H = ADDR >> 16 + ADDR[15]，其中ADDR[15]表示ADDR的第15位的值(这是ADDR_L的符号位.
Mike Spivey correctly noted that the 16-bit offset immediate is sign-extended before being added to the base register.
This calls for a correction of the value I called ADDR_H in case ADDR_L turns out to be negative when interpreted as a 16-bit two's complement number.
If this turns out to be true, ADDR_H must be incremented.
The general formula for ADDR_H can be corrected to ADDR_H = ADDR >> 16 + ADDR[15] where ADDR[15] denotes the value of bit 15 of ADDR (which is the sign bit of ADDR_L.
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