Description
PART I (10 POINTS) In the class lecture, we have compiled the following C code: while (save[i] == k) i += 1; into the following assembly language: Loop: sll $t1, $s3, 2 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, 1 j Loop Exit: . . . Assume the memory address for the first instruction (Loop: …) is 0x00400020. Now, you are expected to assemble the program (except the last line of the assembly code) into machine code. Your solution should contain three columns: the first column being the memory address of the instruction, the second column being the binary representation of the machine code, and the third column being the hexadecimal representation of the machine code.
Memory Address Machine Code (Bin) Machine Code (Hex) 0x00400020 000000, 00000, 10011, 01001, 00010, 000000 0x00134880
PART II (9 POINTS) For the following C code segment, write a code segment in MIPS assembly language to do the same thing. Assume i is in $s0, x is in $s1, and y is in $s2. Don’t forget to comment you code. for (i=0; i
bge $t5, $t3, L If ($t5 >= $t3) go to L
lw $t5, big(t2) $t5 = Memory[$t2 + big] PART IV (10 POINTS) In the class lecture, we have compiled the following C code: int fact (int n) { if (n < 1) return 1; else return (n * fact (n-1)); } into the following assembly code: fact: addi $sp, $sp, -8 #adjust stack pointer sw $ra, 4($sp) #save return address sw $a0, 0($sp) #save argument n slti $t0, $a0, 1 #test for n < 1
beq $t0, $zero, L1 #if n >=1, go to L1 addi $v0, $zero, 1 #else return 1 in $v0 addi $sp, $sp, 8 #adjust stack pointer jr $ra #return to caller L1: addi $a0, $a0, -1 #n >=1, so decrement n jal fact #call fact with (n-1) #this is where fact returns bk_f: lw $a0, 0($sp) #restore argument n lw $ra, 4($sp) #restore return address addi $sp, $sp, 8 #adjust stack pointer mul $v0, $a0, $v0 #$v0 = n * fact(n-1) jr $ra #return to caller Assume the memory address for the first instruction (fact: …) is 0x00400020. Now, you are expected to assemble the program (SKIP the instruction: mul $v0, $a0, $v0) into machine code. Your solution should contain three columns: the first column being the memory address of the instruction, the second column being the binary representation of the machine code, and the third column being the hexadecimal representation of the machine code
