• Byte Memory Array Loading (Basic)
• Hand Assembly (Intermediate)

50.002 Computation Structures
Information Systems Technology and Design
Singapore University of Technology and Design

Assemblers and Compilers

Each topic’s questions are grouped into three categories: basic, intermediate, and challenging. You are recommended to do all basic problem set before advancing further.

The amount of practice problems in this set is smaller than usual because the topics learned this week is mainly to set up the knowledge required for the next topic. You will have more practice hand-assembling C instructions during our lab.

Byte Memory Array Loading (Basic)

The memory is loaded as a byte array {14, 00, 3F, 60, 18, 00, 5F, 60, 00, 10, 01, 80, 1C, 00, 1F, 64, 00, 00, 00, 00, 02, 00, 00, 00, 04, 00, 00, 00, 00, 00, 00, 00} before starting execution.

For example, this means that MEM[0] = 0x14 and MEM[2] = 0x3F and so on.

What is the WORD at memory address 0xC? And what is the word at memory address 0x14?

Hand Assembly (Intermediate)

This beta assembly code provided below counts the number of “bulls” in two sets of 4-integer numbers, each valued between 0-9. A “bull” happens when we have the same number in the same position. This code will be used for the next 2 questions as well.

For example, the following set has 2 “bulls”, because there’s a match in the last digit “0” and in the largest digit “8”.

set_1 = 8190
set_2 = 8920

And this set has 4 bulls because all 4 numbers and 4 positions matches:

set_1 = 1234
set_2 = 1234

We can represent each set in 32-bit, especially the lower 16-bit. Each digit takes up 4-bit of value. For example:

set_1 = 0x2298
set_2 = 0x5678

Here’s the full code.

.include beta.uasm
 
BR(start)
 
.=0x100 | address 256
start:
	CMOVE(0,R1)		| set initial val of var bulls = 0
	CMOVE(0xF,R8)		| set initial val of var mask = 0xF
	CMOVE(0,R3)		| set initial val of var i = 0
	CMOVE(0x0118, R11)	| number set 1 
	CMOVE(0x1218, R12)	| number set 2
 
forbull:
	AND(R11,R8,R9)		| a&mask
	AND(R12,R8,R10)		| b&mask
	CMPEQ(R9,R10,R9)	| if R9==R10 bitwise, R9 = 1
	BEQ(R9, endifbull, R30)	| if R9==0, go to the endifbull
	ADDC(R1,1,R1)		| bull = bull+1
	OR(R11,R8,R11)		| a = a|mask
	OR(R12,R8,R12)		| b = b|mask
 
endifbull:
	SHLC(R8,4,R8)		| mask = mask<<4
	ADDC(R3,1,R3)		| i = i+1
	
endforbull:
	CMPLTC(R3,4,R9)		| if i<4, R9 is 1
	BT(R9, forbull)		| if R9 is 1, go to the forbull loop
 
end:
	LD(R31, answer_key, R2)	| load answer key
	CMPEQ(R2, R1, R0)	| check R2 == R1
    HALT()
 
.=0x200 | address 512
answer_key:
LONG(2)

Hand-assemble certain instructions by providing the higher and lower 16-bit of the machine code for these selected instructions. Present your answer in hexadecimal.

1. AND(R11, R8, R9)
   Show Answer

   Higher 16-bit: `0xA12B`, Lower 16-bit: `0x4000`.

   
   

2. BT(R9, forbull)
   Show Answer

   Higher 16-bit: `0x7BE9`, Lower 16-bit: `0xFFF5`.

   
   

3. LD(R31, answer_key, R2)
   Show Answer

   Higher 16-bit: `0x605F`, Lower 16-bit: `0x0020`.

   
   

What is the content of Reg[R0] when the CPU halts?

The equivalent C code for the above Beta assembly code is shown below, but it is incomplete. Fill in the blanks accordingly without any trailing or leading spaces.

int main(void) {
    int bulls = 0;
    int mask = 0x[Blank 1];
    int i = 0;
 
    int a = 0x0118;
    int b = 0x1218;
 
    while (i < [Blank 2]) {
        int x = a & [Blank 3];
        int [Blank 4] = b & mask;
 
        if ([Blank 5] == y) {
            bulls = [Blank 6] + 1;
            a = [Blank 7] | mask;
            b = b | [Blank 8];
        }
 
        mask = mask << [Blank 9];
        i = i + 1;
    }
 
    int answer_key = [Blank 10];
    int result = (answer_key == [Blank 11]);
 
    return [Blank 12];
}

int main(void) {
    int bulls = 0;
    int mask = 0xF; 
    int i = 0;
 
    int a = 0x0118;
    int b = 0x1218;
 
    while (i < 4) {
        int x = a & mask;
        int y = b & mask;
 
        if (x == y) {
            bulls = bulls + 1;
            a = a | mask;
            b = b | [mask];
        }
 
        mask = mask << 4;
        i = i + 1;
    }
 
    int answer_key = 2;
    int result = (answer_key == bulls);
 
    return result;
}