问题描述
我在LC3机器上编写一个汇编程序。
I am writing an assembly program on the LC3 machine.
我的汇编程序是乘以R2和R3和结果保存在R1的LC3程序。
My assembly program is an LC3 program that multiplies R2 and R3 and stores the result in R1.
下面是我的源$ C $ c（带评论）
Here is my source code(with comments)
;Sets pc to this address at start of program
.ORIG x3000
;R1 will store the result lets clear it(ANd with 0)
AND R1,R1,x0
;R2 will be multiplied by R3, let's clear both of them
AND R2,R2,x0
AND R3,R3,x0
;Test case 4 * 3 = 12;
ADD R2,R2,4
ADD R3,R3,3
;Add to increment zone
LOOP Add R1,R1,R2;
;Decrement the counter, in this case the 3 or R3
ADD R3,R3,x-1
BrP LOOP
HALT
.END
我的测试用例乘以4 * 3的结果应该是12和应当被存储在R1中。然而，当我运行这个程序在我LC3模拟器，这是我得到的输出
My test case is multiplying 4 * 3. The result should be 12 and that should be stored in R1. However when I run this program in my LC3 simulator, this is what i get for the output
R3拥有正确的值时结束，但R1拥有-1 ....有谁看到一个问题，我的code？我确信在开始清除R1和继续增加R3至R1和存储结果到R1，而在这种情况下，计数器，R3或3是大于零。
R3 holds the correct value at the end but R1 holds -1.... Does anyone see an issue with my code? I made sure to clear R1 at the beginning and to keep adding R3 to R1 and storing the result to R1 while the counter, R3, or 3 in this case is greater than zero.
推荐答案
暂停只是一个伪指令为用于暂停机器TRAP指令。
HALT is just a "pseudo-instruction" for a TRAP instruction used to halt the machine.
您可以这样写：
TRAP x25 ;HALT the machine
但这种方式，你需要记住的陷阱向量的位置，在这种情况下 X25 。因此，最好是只使用暂停代替。
But in this way you need to remember the position in the TRAP vector, in this case x25. So is better to just use HALT instead.
其他常见陷阱也有pseduo指令：在， OUT 等
Others common TRAPs also have pseduo-instructions: IN, OUT, etc.
我asume要存储结果的地方。你可以这样做：
I asume you want to store your results somewhere. You could do something like:
;Sets pc to this address at start of program
.ORIG x3000
;R1 will store the result lets clear it(ANd with 0)
AND R1,R1,x0
;R2 will be multiplied by R3, let's clear both of them
AND R2,R2,x0
AND R3,R3,x0
;Test case 4 * 3 = 12;
ADD R2,R2,4
ADD R3,R3,3
;Add to increment zone
LOOP Add R1,R1,R2;
;Decrement the counter, in this case the 3 or R3
ADD R3,R3,x-1
BrP LOOP
ST R1, Result ;STORE R1 at Result
HALT
Result .FILL x0000 ;this will be x000C=12h after execution
.END
---------------------编辑---------------------- ----
关于你的最后一个问题（在评论）
About you last question (in comments):
如果HALT停止我的程序，怎么会Reslt .fill伪指令X0000运行
然后呢？
这更是一个关于汇编是如何工作的问题。
This is more a question about how assemblers works.
答案是因为：组装时间 = 执行时间
指令是在大会的时间考虑。
Directives are considered at Assembly Time.
事实上，组装时间由两关：
In fact, Assembly Time is composed in two passes:
- 解析符号创建一个符号表
- 转换的说明真正可执行/机code使用符号表中。
这是实现装配一个很常见的方式，和LC3汇编器是不是例外。
This is a very common way to implement assemblers, and the LC3 assembler is not the exception.
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