In the recursive factorial code given below, what happens if the instruction at line number 0x00400040 (addi $sp,$sp,8) is deleted by mistake from the following code? Ox00400028 fact: addi 0x0040002C 0x00400030 0x00400034 0x00400038 0x0040003C 0x00400040 0x00400044 0x00400048 L1: 0x0040004C 0x00400050 0x00400054 0x00400058 0x0040005C 0x00400060 SW SW siti beq addi addi jr addi jal Iw Iw $sp,$sp, -8 $ra, 4($sp) $a0, 0($sp) $t0, $a0, 1 $t0, $zero, L1 $v0, $zero, 1 $sp,$sp, 8 $ra $a0, $a0, -1 fact $a0, 0($sp) $ra, 4($sp) addi $sp,$sp,8 mult $v0, $a0,$v0 jr $ra Possible answers are as follows: a) Program enters an infinite loop and remains in the loop with the correct value stored in the vo register b) Program enters an infinite loop and remains in the loop with incorrect return value in the vo register c) Program enters an infinite loop and crashes due to memory access that exceeds the stack space

13. Given the following Java loop, write the equivalent MIPS instructions. for (int i = 5; i < 75; i =i+ 10) { for (int j = 3; j< 40; j=j+ 13) { }

The codes below are disassembled from a program. Write an invalid target for the “jmp” instruction such that the code can be considered as anti-disassembly. Jmp ??? loc_2: call 15FF2A1h OR [ecx], d1 inc eax jmp short loc_3 db 0E8h   Please just give me a simple answer

For example, instruction like this.   0011 0000 0000 0000 ; start at x30000001 001 001 1 00000 ;x3000 set N CC if R1 < 00000 100 0 0000 0001 ;x3001 if N goto x3003 R2+=50000 111 0 0000 0001 ;x3002 Not N goto x30040001 010 010 1 00101 ;x3003 R2 += 5;x3004 done

There are three single precision floating point numbers, numA, numB and numC. Write a piece of code in MIPS assembly language  that compares  two numbers numA and numb, if numB  is greater than numA, add them and save the result in numC. Otherwise save 0 in numC.

This is Computer Architecture! Instructions: Create a MIPS program that demonstrates that the associative law fails in addition for floating point numbers (single or double precision). You only need to demonstrate it for single precision. Remember the associative law is a + (b + c) = (a + b) + c. The program’s output should look something like the following where the xxx’s are the numbers you chose. The resulting numbers may be different than mine depending on your choice of a, b, and c. Using a = xxx, b = xxx, and c = xxx a + (b + c) = 0 (a + b) + c = 1 The key is to have two of the number large (one positively and one negatively but equal in magnitude) floating point numbers and the third floating point number very small in comparison. Please don't forget to include comments in this code

What is the execution result for each of the following instructions : - 1. LHLD 3003 ( [3OO2]=9OH,[30O4]=99H , [3OO1]=1OH,[3OO3]=43H)2.CMA ( If A= 9H ) 3. ADI A, 15H ( A=O1H) 4. INB (If B=34 H) 5. LXI * H 50 H

After the SHR instruction is executed in the code sequence below, what is the final value for AL? MOV AL, 1010_1111b SHR AL, 2 The answer is 0010 0111. а. The answer is 0010_1011. Ob. The answer is 0011_0011. Oc. The answer is 0010_0011. Od.

Which of the following statements about the "count leading zeros" operation is false? It cannot be accomplished in less than O(n) time for n bits It is useful for normalization in floating-point arithmetic An efficient implementation could be written as a recursive Verilog modula

Write a Little Man Computer program that accepts five random numbers from the user and displays them in ascending order. Attached is the Little Man Mnemonic Instruction Codes: