如何识别n，i，x，e位设置(SIC/XE)？

Question

我有一个系统软件类的项目，我们必须生成一个子例程，将一行源代码分解为4个组件: label、op代码、operand1和operand2，并标识n、i、x和e位设置。我想弄清楚这点有点问题。预先感谢您的帮助

下面是一些示例:以下4个源行示例的组件和位如下：

1. 示例+LDA组件，X .MAIN

标签：“示例”op代码："LDA“操作数1：”填充“操作数2："X"nixbpe: 111?1

• RSUB无OP

标签：“op代码："RSUB”操作数1："“NO”操作数2："OP“

nixbpe: 110?0

• CMT字节@C‘’RESERVED块，XYZ备用

标签："CMT“op代码：”字节“操作数1：”C‘’RESERVED块‘“操作数2："XYZ”<注意:只要'X’是第二个operand>的第一个字符，就会设置nixbpe中的x(索引)位。

nixbpe: 101?0

• RMO A，X

标签：“op代码："RMO”操作数1："A“操作数2："X”

：11l??0

标记为"?“保持从调用程序接收的状态；op代码的前缀确定e位；operand1的前缀确定n位和i位；operand2的第一个字符决定x位。B位和p位在其他地方设置。在这个例程中不需要进行语义检查(特别是，上面对RSUB语句的两种解释都是可以的)。组件的默认值是空字符串。默认的"ni??pe“位是"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)***

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