Question

我的系统软件类有一个项目，我们必须生成一个子程序，将一行源分解为4个组件：标签，操作码，操作数1和操作数2，并识别n，i，x和e位设置。我试图找出nixbpe位有问题。提前感谢您的帮助

以下是一些例子：以下4个源代码行示例的组件和位如下：

示例+ LDA STUFF，X .MAIN

标签：“示例” 操作码：“LDA” 操作数1：“STUFF” 操作数2：“X”

nixbpe：111 ?? 1
RSUB NO OP

标签：“” 操作码：“RSUB” 操作数1：“否” 操作数2：“OP”

nixbpe：110 ?? 0
CMT BYTE @ C'RESERVED BLOCK'，XYZ SET ASIDE

标签：“CMT” 操作码：“BYTE” 操作数1：“C'RESERVED BLOCK'” 操作数2：“XYZ”＆lt;注意：只要'X'是第二个操作数的第一个字符＆gt;就会设置nixbpe中的x（索引）位。

nixbpe：101 ?? 0
RMO A，X

标签：“” 操作码：“RMO” 操作数1：“A” 操作数2：“X”

nixbpe：11l ?? 0

nixbpe位标记为“？”保持从呼叫程序收到;操作码的前缀决定了e位; operand1的前缀确定n和i位; operand2的第一个字符确定x位。 b和p位在别处设置。在此例程中不需要进行语义检查（因此，特别是上面对RSUB语句的两种解释都可以）。组件的默认值是空字符串。 “ni ?? pe”位的默认值为“00 ?? 00”。

Answer 1

***寻址标志位描述

Mode        n i x b p e
-------------------------------------
Direct      1 1 0 0 0 0   12 bit displacement is target address
            1 1 0 0 0 1   20 bit address is target address
            1 1 0 0 1 0   12 bit 2's complement displacement from PC (PC relative)
            1 1 0 1 0 0   12 bit base unsigned displacement forward from B (base displacement)
            1 1 1 0 0 0   index register X added to direct address to get target address
            1 1 1 0 0 1   index register X added to direct address to get target address
            1 1 1 0 1 0   index register X added to PC relative computation to get target address
            1 1 1 1 0 0   index register X added to base displacement computation to get target address
            0 0 0 - - -   simple SIC instruction, last 15 bits are the address
            0 0 1 - - -   index register X added to direct address to get target address
Indirect    1 0 0 0 0 0   Computed memory address contains the target address
            1 0 0 0 0 1   Computed memory address contains the target address
            1 0 0 0 1 0   Computed memory address contains the target address
            1 0 0 1 0 0   Computed memory address contains the target address
Immediate   0 1 0 0 0 0   Computed memory address is the operand (target address is the instruction)
            0 1 0 0 0 1   Computed memory address is the operand (target address is the instruction)
            0 1 0 0 1 0   Computed memory address is the operand (target address is the instruction)
            0 1 0 1 0 0   Computed memory address is the operand (target address is the instruction)***

有关详细信息，请参阅http://www.unf.edu/~cwinton/html/cop3601/supplements/sicsim.docL.html

如何识别n，i，x和e位设置（SIC / XE）？

1 个答案: