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A Simple Machine Language
In this appendix we present a simple but representative machine language. We begin by explaining the architecture of the machine itself.
The Machine’s Architecture
The machine has 16 general-purpose registers numbered 0 through F (in hexadec- imal). Each register is one byte (eight bits) long. For identifying registers within instructions, each register is assigned the unique four-bit pattern that represents its register number. Thus register 0 is identified by 0000 (hexadecimal 0), and register 4 is identified by 0100 (hexadecimal 4).
There are 256 cells in the machine’s main memory. Each cell is assigned a unique address consisting of an integer in the range of 0 to 255. An address can therefore be represented by a pattern of eight bits ranging from 00000000 to 11111111 (or a hexadecimal value in the range of 00 to FF).
Floating-point values are assumed to be stored in an eight-bit format discussed in Section 1.7 and summarized in Figure 1.24.
The Machine’s Language
Each machine instruction is two bytes long. The first 4 bits provide the op-code;
the last 12 bits make up the operand field. The table that follows lists the instruc- tions in hexadecimal notation together with a short description of each. The let- ters R, S, and T are used in place of hexadecimal digits in those fields representing a register identifier that varies depending on the particular application of the instruction. The letters X and Y are used in lieu of hexadecimal digits in variable fields not representing a register.
C
a p p e n d i x
Op-code Operand Description
1 RXY LOAD the register R with the bit pattern found in the memory cell whose address is XY.
Example: 14A3 would cause the contents of the memory cell located at address A3 to be placed in register 4.
2 RXY LOAD the register R with the bit pattern XY.
Example: 20A3 would cause the value A3 to be placed in register 0.
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582 Appendixes
Op-code Operand Description
3 RXY STORE the bit pattern found in register R in the memory cell whose address is XY.
Example: 35B1 would cause the contents of register 5 to be placed in the memory cell whose address is B1.
4 0RS MOVE the bit pattern found in register R to register S.
Example: 40A4 would cause the contents of register A to be copied into register 4.
5 RST ADD the bit patterns in registers S and T as though they were two’s complement representations and leave the result in register R.
Example: 5726 would cause the binary values in registers 2 and 6 to be added and the sum placed in register 7.
6 RST ADD the bit patterns in registers S and T as though they represented values in floating-point notation and leave the floating-point result in register R.
Example: 634E would cause the values in registers 4 and E to be added as floating-point values and the result to be placed in register 3.
7 RST OR the bit patterns in registers S and T and place the result in register R.
Example: 7CB4 would cause the result of ORing the contents of registers B and 4 to be placed in register C.
8 RST AND the bit patterns in registers S and T and place the result in register R.
Example: 8045 would cause the result of ANDing the contents of registers 4 and 5 to be placed in register 0.
9 RST EXCLUSIVE OR the bit patterns in registers S and T and place the result in register R.
Example: 95F3 would cause the result of EXCLUSIVE ORing the contents of registers F and 3 to be placed in register 5.
A R0X ROTATE the bit pattern in register R one bit to the right X times.
Each time place the bit that started at the low-order end at the high-order end.
Example: A403 would cause the contents of register 4 to be rotated 3 bits to the right in a circular fashion.
B RXY JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of execution. (The jump is implemented by copying XY into the program counter during the execute phase.) Example: B43C would first compare the contents of register 4 with the contents of register 0. If the two were equal, the pattern 3C would be placed in the program counter so that the next instruction executed would be the one located at that memory address. Otherwise, nothing would be done and program execution would continue in its normal sequence.
C 000 HALT execution.
Example: C000 would cause program execution to stop.
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