The content of a part of MIPS memory and the contents of three registers are provided below. Answer the following questions?
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- The content of registers and memory are summarized as follow: R0=0x24682244 R1=0x00446688 R2=0x00118822 R3=0X00000077 R4=0x4000000C R5=0x00000008 R6=0x00000055 R7=0x40000000 R8=0x40000002 (0x40000000)= 12 84 56 78 90 AB CD EF 11 22 33 44 55 66 77 88 Evaluate the following program and obtain the new content of the registers R0 – R7, and new content of memory starting from address 0x40000000 until 0x4000000F ENTRY MOVS R0,#25 LDRB R1,[R7] LDRH R2,[R7,#0x2] LDR R3,[R7] LDRSB R4,[R7,R5] LDRSH R5,[R8,R5] STR R6,[R7],#4 STRB R7,[R8,#1]! STM R7,{R1,R5-R6} STRH R0,[R8,#5] ENDA 74LS139 2-to-4 code is used to partition the memory space of an MC68008 into 4 quadrant, ROM, RAM, PER (for Synchronous devices) and the 4th is reserved for future provision. Write the address range of the RAM. And compute the addresses per location in ROM if is 64 KBGiven the memory content below, fill in the table below the registers values. Initial values for PC=20A, AC=FFF8. (all values are in Hexadecimal) address Content 208 0209 209 000A 20A 7008 20B 1109 20C 9208
- Question 2: Write the MIPS code for the given Hexadecimal Machine Code that starts at memory address 0x40000. Line1: 0x0080082A Line2: 0x14200002 Line3: 0x2084FFFF Line4: 0x08010005 Line5: 0x20840001 Line6: 0x1480FFFA Line7: 0x03E00008 Line8: 0x0C010000Translate the following MIPS machine code into MIPS assembly language.0x2010000a0x341100050x012ac0220x001840820x030f9024Given the binary format of an instruction as follows: 0000 1101 0110 1100 1010 1011 0001 0111 a. What assembly instruction does this correspond to? b. If the PC + 4 = 0XA0BC32F4, what is the target address? c. How many instructions forwards or backwards is that? d. How many bytes is that in decimal?
- 0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.The following are the 32-bit hexadecimal numbers that represent the machine code of six different MIPS instructions. For each one of these instructions determine the source and destination registers, if any, represented by their name (e.g., Șt5) not their code number. Explain your answer by finding the assembly code of each instruction. а. Өхe251902е b. exe0129080 c. exae4a0000 d. ex22100001 e. ex1608fff3 f. ex8e4a0005Part 3: Putting it all together Consider the situation where the EAX and EBX registers are storing the values listed below. EAX: 01100001 10101110 00101111 11111111 EBX: 11101010 00000000 00000010 10101101 We want to add the values of EAX and EBX together. When we perform this addition, the result is stored EBX, overwriting the value that was already in there. What is the result of the addition stored in EBX? Give your result in binary. What value is stored in BL? Give your result in hexadecimal. What flags are set after the operation is complete?
- What is the effective address that is targeted by the store instruction whose code word in binary is:101011 01000 10001 1111 1111 1111 1000Assume [$t0]=0x400CThree Instructions and One Value Here are three segments of a program. After the LC-3 executes the three instructions starting at address x4810 a value is put into register R6. x4010 1110110000101111 x4011 011011e110111111 x4012 011011e110111111 X403F 01100ee010110011 x4040 01100ee010110011 x4041 01100e001011e010 X60B2 11001e1011111110 X60B3 1011111010101011 X60B4 1011111010101101 The hex (with an x) value in register R6:Given the RISC instruction xorcc %i1, 885, %g4 Answer the following questions regarding the 32-bits of this instruction (all answers should be binary one's and zero's and should NOT contain any spaces): a) What are the two most significant bits of this instruction? 10 b) What are the five destination register bits? c) What are the six operation bits? 010011 d) What are the five first operand bits? e) What is the value of bit 13? f) What are the second operand bits stored in instruction bits 12 down to 0? (must be appropriate size in binary)