CS 366 - Mythsim Microprogram Assignment

Solution submitted by one of the groups in the class

Note: in the Mythsim simulator, the c_out status line actually contains the inverted value of the carry out instead of the straight (non-complimented) value of the carry out.

Part 0 is already given to you
Part 1 is due on Friday 10/18/2002
Part 2 is due on ~~Friday 11/1/2002~~ Monday 11/4/2002. The turnin command will be turned off on Monday after Prof. Troy returns to his office from class (arouind 3:15 pm).

You are to work in groups of 2 or 3 students for this assignment. Working in groups is required for the assignment. For Part 1, the work you turn in must include the information from Part 0. For Part 2, the work you turn in must include the information from Part 0 and Part 1.

For part 1 of this assignment, your group will need to write a microprogram for the Mythsim machine for all opcodes specified below for parts 0 or 1. You will need to turn this program in electronically using the turnin command and you will to bring a hard copy of your program to class on the day it is due. For part 1, your program does not have to run on the Mythsim simulator (version 8), but you are highly encouraged to get it running on it. Here is a working Mythsim program for part 0.

For part 2, you will have to electronically turn in (via the UNIX turnin command on the CS machines) a microprogram that runs on the the MythSim simulator (release 8). This microprogram must run all opcodes as specified in this write up. Here is a working Mythsim program for part 0.

Here are two mem files, that show the machine code for the assembly language instructions for two programs. The first program just lists all of the operations for part0. The second program lists the operations to perform an assembly langauge equivelent to x = y + z.

The grading for part 2 will be done as follows:

90% of the grade will be for writing working microcode for all of the opcodes specified.
5% of the grade will be for a ranking of how small your microcode is compared to the rest of the class. Note microprograms that do not get all of the opcodes working will not be elligible for this portion of the assignment.
5% of the grade will be for a ranking of how fast your microcode is compared to the rest of the class when running certain test program(s) designed by the instructor and TA. Note microprograms that do not get all of the opcodes working will not be elligible for this portion of the assignment.

Assignment
Part Opcode
number Name Description
0 0 no-op This operation does nothing
0 1 add Ri ← Rj + Rk
This operation will add the values from registers Rj and Rk and store the result in resiter Ri.
0 2 set register Ri ← Const8
This operation will store in register Ri the constant value given in the last 8 bits of the instruction register.
0 3 Branch if zero if (Rj == 0) then PC ← PR + Const4
This instruction will check if the value in register Rj is zero. If so, it will add the 4 bit constant value from the instruction register to the PC.
0 4 move Ri ← Rj
This instruction will take the value in register Rj and store it in register Ri.
0 5 Store Mem[Rj] ← Rk
This instruction will take a value in register Rk and store it in the computer's memory. The memory address is given by the value in register Rj.
0 6 Load Ri ← Mem[Rj]
This instruction will take a value from the computer's memory and store it in register Ri. The memory address is given by the value in register Rj.
0 7 Subtraction Ri ← Rj - Rk
This operation will subtract from the value in register Rj the value in register Rk and store the result in resiter Ri.
1 8 Bitwise OR Ri ← Rj | Rk
This operation will perform a BITWISE OR (each bit position is ORed independently of the other bit positions). It will or each bit of the value in register Rj with the value in register Rk and store the result in register Ri.
1 9 Bitwise AND Ri ← Rj & Rk
This operation will perform a BITWISE AND (each bit position is ANDed independently of the other bit positions). It will and each bit of the value in register Rj with the value in register Rk and store the result in register Ri.
1 10 Bitwise NOT Ri ← ¬ Rj
This operation will perform a BITWISE NOT (each bit position is NOTed independently of the other bit positions). It will not (invert) each bit of the value in register Rj and store the result in register Ri.
2 11 Bitwise XOR Ri ← Rj \305 Rk
This operation will perform a BITWISE XOR (each bit position is XORed independent of the other bit positions). It will XOR each bit of the of the value in register Rj with the corresponding bit of the valus in register Rk and store the result in register Ri.
2 12 Bitwise NOR Ri ← ¬ (Rj | Rk)
This operation will perform a BITWISE NOR (each bit position is NORed independent of the other bit positions). It will NOR each bit of the of the value in register Rj with the corresponding bit of the valus in register Rk and store the result in register Ri.
1 13 Increment Ri ← Rj + Const4
This operation will add to the value in regsiter Rj the sign extented value contained in the Const4 field of the instruction register and store this result in register Ri. Note the sign extention of the 4 bit constant value is automatically done by the machine.
1 14 Negation Ri ← -Rj
This operation will change the sign on the value in register Rj and store the result in register Ri.
In effect this operation performs: Ri ← 0 - Rj.
1 15 Logical Left Shift Ri ← Rj<<Rk
This operation will shift the value in register Rj to the left by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros for each bit position shifted. Note that if the value in register Rk is 8 or greater, the result will be all zeros. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2 16 Logical Right Shift Ri ← Rj>>Rk
This operation will shift the value in register Rj to the right by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros for each bit position shifted. Note that if the value in register Rk is 8 or greater, the result will be all zeros. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2 17 Arithmetic Right Shift Ri ← Rj>>Rk
This operation will shift the value in register Rj to the right by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros or ones depending on the sign bit of the original value in Rj. The sign bit is the most significant bit (the leftmost). If the sign bit is zero, the shifted value is filled in with zeros. If the sign bit is one, the shifted value is filled in with ones. Note that if the value in register Rk is 8 or greater, the result will be all zeros or all ones (depending on the sign bit of the original value in Rj. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2 18 Circullar Left Shift Ri ← Rj<<Rk
This operation will circular shift the value in register Rj to the left by the number of bits equal to the value in register Rk. The result will be stored in register Ri. For each position shifted, the value is filled in with a zero if the bit value shifted out was zero and the value is filled in with a one if the value shifted out was one. Note that if the value in register Rk is a multiple of 8, the result will be the original value. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, etc).
1 2 19 Multiplication Ri ← Rj * Rk
This operation will multiply the value in register Rj by the value in register Rk and store the result in register Ri. Only the lower 8 bits of the result is stored in register Ri. Do not worry about overflow.
2 20 Division Ri ← Rj / Rk
This operation will divide the value in register Rj by the value in register Rk and store the integer quotent in register Ri.
2 21 Remainder Ri ← Rj % Rk
This operation will divide the value in register Rj by the value in register Rk and store the integer remainder in register Ri.
2 22 Bit Set Ri ← Rj with the Rk-th bit of Rj set to one
If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2 23 Bit Clear Ri ← Rj with the Rk-th bit of Rj cleared to zero
If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2 24 Bit Invert Ri ← Rj with the Rk-th bit of Rj inverted
If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2 25 Branch if Bit Set if (the Rk-th bit of Rj == 1) then PC ← PC + Const4
If the value in register Rk is outside of the range from 0 to 7, the operation does not branch (the program counter is not changed).
1 26 Branch if equal if (Rj == Rk) then PC ← PC + Const4
If the value in register Rj is equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
1 27 Branch if not equal if (Rj != Rk) then PC ← PC + Const4
If the value in register Rj is not equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2 28 Branch if less than if (Rj < Rk) then PC ← PC + Const4
If the value in register Rj is less than to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2 29 Branch if less than or equal if (Rj <= Rk) then PC ← PC + Const4
If the value in register Rj is less than or equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2 30 Branch if greater than if (Rj > Rk) then PC ← PC + Const4
If the value in register Rj is greater than to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2 31 Branch if greater than or equal if (Rj >= Rk) then PC ← PC + Const4
If the value in register Rj is greater than or equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2 32 Branch if less than zero if (Rj < 0) then PC ← PC + Const4
The the value in register Rj is less than zero, add the 4 bit constant value from the instruction register to the program counter.
2 33 Branch if greater than zero if (Rj > 0) then PC ← PC + Const4
The the value in register Rj is greater than zero, add the 4 bit constant value from the instruction register to the program counter.
2 34 Branch if less than or equal to zero if (Rj <= 0) then PC ← PC + Const4
The the value in register Rj is less than or equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2 35 Branch if greater than or equal to zero if (Rj >= 0) then PC ← PC + Const4
The the value in register Rj is greater than or equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2 36 Branch if not zero if (Rj != 0) then PC ← PC + Const4
The the value in register Rj is not equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2 37 Jump with Constant PC ← Const8
Store in the program counter (register 7) the value in the 8 bit constant field in the instruction.
2 38 Jump with Register PC ← Rj
Store in the program counter (register 7) the value from register Rj.
2 39 Jump to Subroutine Ri ← PC ; PC ← Const8
First, store the current value of the program counter (register 7) in register Ri. Then, store in the program counter (register 7) the value in the 8 bit constant field in the instruction.
2 40 Return from Subroutine PC ← Rj
Store in the program counter (register 7) the value from register Rj.
1 41 Halt Stop exectution
In Mythsim, execution is halted when the machine detects a microinstruction that does a goto to itself. The microinstruction for this is:
opcode[41]: goto opcode[41];
Remember that the instruction has two formats. The first format is:

Opcode
6 bits Ri
2 bits Rj
2 bits Rk
2 bits Const4
4 bits
The second format is:

Opcode
6 bits Ri
2 bits Const8
8 bits

Assignment Part	Opcode number	Name	Description
0	0	no-op	This operation does nothing
0	1	add	Ri ← Rj + Rk This operation will add the values from registers Rj and Rk and store the result in resiter Ri.
0	2	set register	Ri ← Const8 This operation will store in register Ri the constant value given in the last 8 bits of the instruction register.
0	3	Branch if zero	if (Rj == 0) then PC ← PR + Const4 This instruction will check if the value in register Rj is zero. If so, it will add the 4 bit constant value from the instruction register to the PC.
0	4	move	Ri ← Rj This instruction will take the value in register Rj and store it in register Ri.
0	5	Store	Mem[Rj] ← Rk This instruction will take a value in register Rk and store it in the computer's memory. The memory address is given by the value in register Rj.
0	6	Load	Ri ← Mem[Rj] This instruction will take a value from the computer's memory and store it in register Ri. The memory address is given by the value in register Rj.
0	7	Subtraction	Ri ← Rj - Rk This operation will subtract from the value in register Rj the value in register Rk and store the result in resiter Ri.
1	8	Bitwise OR	Ri ← Rj \| Rk This operation will perform a BITWISE OR (each bit position is ORed independently of the other bit positions). It will or each bit of the value in register Rj with the value in register Rk and store the result in register Ri.
1	9	Bitwise AND	Ri ← Rj & Rk This operation will perform a BITWISE AND (each bit position is ANDed independently of the other bit positions). It will and each bit of the value in register Rj with the value in register Rk and store the result in register Ri.
1	10	Bitwise NOT	Ri ← ¬ Rj This operation will perform a BITWISE NOT (each bit position is NOTed independently of the other bit positions). It will not (invert) each bit of the value in register Rj and store the result in register Ri.
2	11	Bitwise XOR	Ri ← Rj \305 Rk This operation will perform a BITWISE XOR (each bit position is XORed independent of the other bit positions). It will XOR each bit of the of the value in register Rj with the corresponding bit of the valus in register Rk and store the result in register Ri.
2	12	Bitwise NOR	Ri ← ¬ (Rj \| Rk) This operation will perform a BITWISE NOR (each bit position is NORed independent of the other bit positions). It will NOR each bit of the of the value in register Rj with the corresponding bit of the valus in register Rk and store the result in register Ri.
1	13	Increment	Ri ← Rj + Const4 This operation will add to the value in regsiter Rj the sign extented value contained in the Const4 field of the instruction register and store this result in register Ri. Note the sign extention of the 4 bit constant value is automatically done by the machine.
1	14	Negation	Ri ← -Rj This operation will change the sign on the value in register Rj and store the result in register Ri. In effect this operation performs: Ri ← 0 - Rj.
1	15	Logical Left Shift	Ri ← Rj<<Rk This operation will shift the value in register Rj to the left by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros for each bit position shifted. Note that if the value in register Rk is 8 or greater, the result will be all zeros. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2	16	Logical Right Shift	Ri ← Rj>>Rk This operation will shift the value in register Rj to the right by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros for each bit position shifted. Note that if the value in register Rk is 8 or greater, the result will be all zeros. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2	17	Arithmetic Right Shift	Ri ← Rj>>Rk This operation will shift the value in register Rj to the right by the number of bits equal to the value in register Rk. The result will be stored in register Ri. The value is filled in with zeros or ones depending on the sign bit of the original value in Rj. The sign bit is the most significant bit (the leftmost). If the sign bit is zero, the shifted value is filled in with zeros. If the sign bit is one, the shifted value is filled in with ones. Note that if the value in register Rk is 8 or greater, the result will be all zeros or all ones (depending on the sign bit of the original value in Rj. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, set the value to zero, etc).
2	18	Circullar Left Shift	Ri ← Rj<<Rk This operation will circular shift the value in register Rj to the left by the number of bits equal to the value in register Rk. The result will be stored in register Ri. For each position shifted, the value is filled in with a zero if the bit value shifted out was zero and the value is filled in with a one if the value shifted out was one. Note that if the value in register Rk is a multiple of 8, the result will be the original value. The value in Rk is assumed to be greater than or equal to zero. If the value in Rk is negative, do something reasonable (leave the value unshifted, etc).
1 2	19	Multiplication	Ri ← Rj * Rk This operation will multiply the value in register Rj by the value in register Rk and store the result in register Ri. Only the lower 8 bits of the result is stored in register Ri. Do not worry about overflow.
2	20	Division	Ri ← Rj / Rk This operation will divide the value in register Rj by the value in register Rk and store the integer quotent in register Ri.
2	21	Remainder	Ri ← Rj % Rk This operation will divide the value in register Rj by the value in register Rk and store the integer remainder in register Ri.
2	22	Bit Set	Ri ← Rj with the Rk-th bit of Rj set to one If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2	23	Bit Clear	Ri ← Rj with the Rk-th bit of Rj cleared to zero If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2	24	Bit Invert	Ri ← Rj with the Rk-th bit of Rj inverted If the value in register Rk is outside of the range from 0 to 7, the operation is the same as a move operations (see opcode 4). In other words, when the value in register Rk is outside of the range from 0 to 7, the value in Rj is stored as is into register Ri.
2	25	Branch if Bit Set	if (the Rk-th bit of Rj == 1) then PC ← PC + Const4 If the value in register Rk is outside of the range from 0 to 7, the operation does not branch (the program counter is not changed).
1	26	Branch if equal	if (Rj == Rk) then PC ← PC + Const4 If the value in register Rj is equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
1	27	Branch if not equal	if (Rj != Rk) then PC ← PC + Const4 If the value in register Rj is not equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2	28	Branch if less than	if (Rj < Rk) then PC ← PC + Const4 If the value in register Rj is less than to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2	29	Branch if less than or equal	if (Rj <= Rk) then PC ← PC + Const4 If the value in register Rj is less than or equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2	30	Branch if greater than	if (Rj > Rk) then PC ← PC + Const4 If the value in register Rj is greater than to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2	31	Branch if greater than or equal	if (Rj >= Rk) then PC ← PC + Const4 If the value in register Rj is greater than or equal to the value in register Rk, it will add the 4 bit constant value from the instruction register to the program counter.
2	32	Branch if less than zero	if (Rj < 0) then PC ← PC + Const4 The the value in register Rj is less than zero, add the 4 bit constant value from the instruction register to the program counter.
2	33	Branch if greater than zero	if (Rj > 0) then PC ← PC + Const4 The the value in register Rj is greater than zero, add the 4 bit constant value from the instruction register to the program counter.
2	34	Branch if less than or equal to zero	if (Rj <= 0) then PC ← PC + Const4 The the value in register Rj is less than or equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2	35	Branch if greater than or equal to zero	if (Rj >= 0) then PC ← PC + Const4 The the value in register Rj is greater than or equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2	36	Branch if not zero	if (Rj != 0) then PC ← PC + Const4 The the value in register Rj is not equal to zero, add the 4 bit constant value from the instruction register to the program counter.
2	37	Jump with Constant	PC ← Const8 Store in the program counter (register 7) the value in the 8 bit constant field in the instruction.
2	38	Jump with Register	PC ← Rj Store in the program counter (register 7) the value from register Rj.
2	39	Jump to Subroutine	Ri ← PC ; PC ← Const8 First, store the current value of the program counter (register 7) in register Ri. Then, store in the program counter (register 7) the value in the 8 bit constant field in the instruction.
2	40	Return from Subroutine	PC ← Rj Store in the program counter (register 7) the value from register Rj.
1	41	Halt	Stop exectution In Mythsim, execution is halted when the machine detects a microinstruction that does a goto to itself. The microinstruction for this is: opcode[41]: goto opcode[41];

Below is a table showing how to check for a comparison between to values. Remember, to compare A with B, look at the result of A - B. The comparision uses the following pieces of information:

z - if the result of the subtraction is zero (z is 1 if the result is zero, z is 0 if the result is not zero)
s - the sign bit of the result (this is the m7 status line in Mythsim)
v - if the result of the subtraction caused overflow (v is 1 if overflowed occurred, v is 0 if overflow did not occur)

Comparison Operation	equivalent to
A > B	!z && (s == v)
A >= B	s == v
A < B	s XOR v
A >= B	z \|\| (s XOR v)
A == B	z
A != B	!z