编写返回数字是否为奇数的非常基本的 SPARC 汇编例程时遇到问题
我正在编写一个名为 isOdd 的小型汇编例程，顾名思义，如果传递的整数为奇数，则通过从 % 操作返回 1 来返回。
到目前为止，这是我的代码：
Function prototype: int isOdd( long num )
isOdd:
save %sp, -96, %sp ! Save caller's window
mov %i0, %o0 ! Parameter num goes to %o0
mov 2, %l0 ! 2 goes to local register
call .rem ! Call modulus subroutine
nop
mov %o0, %l0 ! moves the result of the subroutine
! to output register o0
ret
restore
但是，我没有得到好的输出；事实上，它似乎只是返回我传递给 num 的任何值，而不是实际执行模数运算。
谷歌尚未证明对这样一个基本问题有帮助。这是我的第一个汇编代码，所以我对“寄存器”的概念非常不熟悉，我认为将它们混合在一起可能是我的错误所在。
预先感谢您的帮助！
I'm writing a small assembly routine called isOdd, which, as the name implies, returns if the passed integer is odd, by returning 1 from a % operation.
This is my code so far:
Function prototype: int isOdd( long num )
isOdd:
save %sp, -96, %sp ! Save caller's window
mov %i0, %o0 ! Parameter num goes to %o0
mov 2, %l0 ! 2 goes to local register
call .rem ! Call modulus subroutine
nop
mov %o0, %l0 ! moves the result of the subroutine
! to output register o0
ret
restore
However, I don't get good output; in fact, it seems like it is just returning whatever value I pass to num, instead of actually doing the modulus operation.
Google hasn't proved helpful for such a basic question. This is my first assembly code, so I'm pretty unfamiliar with the concept of "registers," and I think mixing them up is where my error may lie.
Thanks in advance for your help!
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有一大堆寄存器，您可以将它们视为块中
8 个。在任意时刻，三个连续的 8 个寄存器块可见为
当前寄存器窗口，标记为
%o0-%o7、%l0-%l7，以及%i0-%i7。 （第四个块有 8 个寄存器，%g0-%g7，它们是全局而不是窗口排列的一部分。）
当您
保存或恢复时，窗口将移动两个个8块。重叠块允许参数和结果传递。寄存器
调用者中名为
%o0-%o7的内容与名为的内容相同被调用者中的
%i0-%i7。 （被调用方中的两个新块是%l0-%l7，这些是该窗口内本地使用的私有内容，并且
%o0-%o7当被调用者想要调用另一个函数时可以使用它。）
用图片更清楚：
您的调用者将
num参数放入%o0（在其窗口中），然后来电你。您
保存以设置一个新窗口，因此您可以在%i0中看到它窗户。
.rem有两个参数。您将它们放在%o0和%o1中（在您的窗口），然后调用它。它将在其
%i0和%i1中看到它们（假设它确实save以设置新窗口）。它将答案放入其%i0中，即你的
%o0。同样，您应该将结果放入
%i0;给你打电话的人都会看到在他们的
%o0中。There are a whole bunch of registers, which you can think of as being in blocks
of 8. At any one time, three consecutive blocks of 8 registers are visible as
the current register window, and are labelled as
%o0-%o7,%l0-%l7, and%i0-%i7. (There is a fourth block of 8 registers,%g0-%g7, which areglobal rather than being a part of the windowing arrangement.)
When you
saveorrestore, the window moves by two blocks of 8. Theoverlapping block allows for parameter and result passing. The registers
which are named
%o0-%o7in the caller are the same ones that are named%i0-%i7in the callee. (The two new blocks in the callee are%l0-%l7,which are private for local use within that window, and
%o0-%o7which thecallee can use when it in turn wants to call another function.)
It's clearer with a picture:
Your caller places the
numargument into%o0(in its window), then callsyou. You
saveto set up a new window, and so you see it in%i0in yourwindow.
.remtakes two parameters. You place these in your%o0and%o1(in yourwindow), then call it. It will see them in its
%i0and%i1(assuming it doesa
saveto set up a new window). It puts the answer in its%i0, which isyour
%o0.Similarly, you should put your result in your
%i0; whoever called you will see itin their
%o0.