LC3 LEA指令和存储的值 [英] LC3 LEA instruction and the value stored

问题描述

我对这个问题感到困惑：什么是存储在寄存器0指令后的值LEA R0，A执行为什么答案是x370C我？估计这应该是A的地址加载到R0？如果是这样，我们如何知道地址？是否有人可以帮忙吗？非常感谢！

  .orig这样X3700
 LEA R0，A
 LDI R2，C R3 LDR，R0，2
 与R 1，R 1，＃0
 在
 ST R0，D
 JSR˚F
 停
˚FLD R1，B
 ADD R1，R1，＃1
 BRP˚F
 RET .fill伪一X1234
 乙.fill伪X370B
 .fill伪ÇX370C
 ð.BLKW 2
 Ë.STRINGZABCD
 摹.fill伪X1234
 。结束
 

解决方案

在code的起源是 x3700 ，你有12条指令，因此地址 A的将 x3700 + X0C = x370C 。正如你猜到了， LEA R0，A 加载 A 的地址复制到 R0 ，所以 R0 将包含 x370C 后第一个指令已被执行。

  .orig这样X3700
3700 LEA R0，A
3701 LDI R2，C
3702 LDR R3，R0 2
         ...
370B RET370℃.fill伪X1234
         ...
 

I am confused by this question: What is the value stored in register 0 after instruction "LEA R0,A" is executed? How come the answer is x370C ? I reckon it is supposed to load the address of A into R0? If so how do we know the address? Can someone please help? Many thanks!

.ORIG X3700
 LEA R0, A
 LDI R2, C LDR R3, R0, 2 
 AND R1, R1, #0 
 IN
 ST R0, D 
 JSR  F 
 HALT
F LD  R1, B
 ADD R1, R1, #1
 BRp F 
 RET

 A .FILL X1234
 B .FILL X370B
 C .FILL X370C
 D .BLKW 2
 E .STRINGZ "ABCD"
 G .FILL X1234
 .END

解决方案

The origin of the code is x3700, and you have 12 instructions, so the address of A will be x3700 + x0C = x370C. As you guessed, LEA R0,A loads the address of A into R0, so R0 will contain x370C after that first instruction has been executed.

        .ORIG X3700
3700     LEA R0, A
3701     LDI R2, C
3702     LDR R3, R0, 2 
         ...
370b     RET

370c     A .FILL X1234
         ...

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