为什么对无符号数字进行MIPS操作会得到带符号的结果? [英] Why do MIPS operations on unsigned numbers give signed results?

问题描述

当我尝试在MIPS中处理无符号整数时，即使我执行的每个操作都是无符号运算，我做的每一个运算的结果都保持有符号(即整数全为2的补码):addu， multu等等...

When I try working on unsigned integers in MIPS, the result of every operation I do remains signed (that is, the integers are all in 2's complement), even though every operation I perform is an unsigned one: addu, multu and so fourth...

当我打印范围在[2^31, 2^32 - 1]内的数字时，我得到它们的溢出"负值，就好像它们是带符号的(我猜是这样).

When I print numbers in the range [2^31, 2^32 - 1] I get their 'overflowed' negative value as if they were signed (I guess they are).

但是，当我尝试这样的事情时:

Though, when I try something like this:

li $v0, 1
li $a0, 2147483648                # or any bigger number
syscall

打印的号码始终为2147483647 (2^31 - 1)

我很困惑...我想念什么?

I'm confused... What am I missing?

PS:我没有包含我的代码，因为它不易读(例如汇编代码)，并且抛开了这个问题， 似乎工作正常.如果有人认为有必要，我会立即将其包括在内！

PS : I haven't included my code as it isn't very readable (such is assembly code) and putting aside this problem, seems to be working fine. If anyone feels it is necessary I shall include it right away!

推荐答案

来自 Wikipedia :

MIPS32指令集指出，作为加减指令的一部分未签名的单词是不正确的. 命令的有符号版本和无符号版本之间的区别不是操作数的符号扩展(或没有符号扩展)，而是控制是在溢出(例如，添加)时执行陷阱还是忽略溢出(添加无符号).这些指令的立即操作数CONST始终进行符号扩展.

The MIPS32 Instruction Set states that the word unsigned as part of Add and Subtract instructions, is a misnomer. The difference between signed and unsigned versions of commands is not a sign extension (or lack thereof) of the operands, but controls whether a trap is executed on overflow (e.g. Add) or an overflow is ignored (Add unsigned). An immediate operand CONST to these instructions is always sign-extended.

从 MIPS指令参考:

所有算术立即数都用符号扩展[...]有符号和无符号指令之间的唯一区别是，有符号指令可以生成溢出异常，而无符号指令则不能.

ALL arithmetic immediate values are sign-extended [...] The only difference between signed and unsigned instructions is that signed instructions can generate an overflow exception and unsigned instructions can not.

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