Ch04.2. Consider the following assembly language code: 10: lw $30, value I1: add $31, $30, $a1 12: add $t1,$t1, Sal 13: lw $t7, ($30) 14: bnez St7, loop 15: add $t1,$t1, $30 16: add $t2, $30, $t4 17: add $t3, $t2, $t5 Consider a pipeline with forwarding, hazard detection, and 1 delay slot for branches. The pipeline is the typical 5-stage IF, ID, EX, MEM, WB MIPS design. For the above code, complete the pipeline diagram below (instructions on the left, cycles on top) for the code. Insert the characters IF, ID, EX, MEM, WB for each instruction in the boxes. Assume that there two levels of bypassing, that the second half of the decode stage performs a read of source registers, and that the first half of the write-back stage writes to the register file. Label all data stalls (Draw an X in the box). Label all data forwards that the forwarding

Ch04.2. Consider the following assembly language code: 10: lw $30, value I1: add $31, $30, $a1 12: add $t1,$t1, Sal 13: lw $t7, ($30) 14: bnez St7, loop 15: add $t1,$t1, $30 16: add $t2, $30, $t4 17: add $t3, $t2, $t5 Consider a pipeline with forwarding, hazard detection, and 1 delay slot for branches. The pipeline is the typical 5-stage IF, ID, EX, MEM, WB MIPS design. For the above code, complete the pipeline diagram below (instructions on the left, cycles on top) for the code. Insert the characters IF, ID, EX, MEM, WB for each instruction in the boxes. Assume that there two levels of bypassing, that the second half of the decode stage performs a read of source registers, and that the first half of the write-back stage writes to the register file. Label all data stalls (Draw an X in the box). Label all data forwards that the forwarding

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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This article examines Windows Service Control Manager.
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In 8085. Please do it in sim8085 compiler
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1.9 Assembly language uses branch instructions to implement loops and selection statements. Explain why a for loop and an if statement in high-level languages represent an improvement on this assembly language technique.
can you show me with actual output console with color like screen shot of the output show me the out put of this assembly language with different in put please INCLUDE Irvine32.inc.stack 4096ExitProcess proto,dwExitCode:dword.data; strings(texts) for promptsprompt_1st_int BYTE "Please enter the first integer: ",0prompt_2nd_int BYTE "Please enter the second integer: ",0prompt_text_color BYTE "Please pick the text color (1=blue, 2=green, 4=red, or any number from 0-15): ",0prompt_proverb BYTE "Please teach me a proverb: ",0response_str BYTE "The proverb I just learned: ", 0; strings(texts) for output labels/descriptionsdec_sum_str BYTE "The sum in DEC is: ",0hex_sum_str BYTE "The sum in HEX is: ",0memory_dump_str BYTE "Memory Dump with operand_1, _2, and sum: ",0; variables definitionsoperand_1 SDWORD ?operand_2 SDWORD ?sum SDWORD ?color_code DWORD ?aString BYTE 80 DUP(0), 0.codemain proc;prompt the 1st number inputmov edx, OFFSET prompt_1st_intcall WriteString; set input text color,…
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THIS IS MY CODE SO FAR: // header #include <stdio.h> // define the IM and DM size #define IM_SIZE 250 #define DM_SIZE 10 // Define the ISA of the Tiny Machine Architecture #define LOAD 1 #define ADD 2 #define STORE 3 #define SUB 4 #define IN 5 #define OUT 6 #define END 7 #define JMP 8 #define SKIPZ 9 // Accurately represented architecture int PC = 0; // PC = Program Counter int IR; // IR = Instruction Register int MAR1, MAR2; // MAR1,MAR2 = Memory Address Registers 1 and 2 int MDR1, MDR2; // MDR1,MDR2 = Memory Data Registers 1 and 2 int A = 0; // A = Accumulator int IM[IM_SIZE]; // IM = Instruction Memory int DM[DM_SIZE]; // DM = Data Memory // opening the file void openProgram(const char* filename) { FILE* file = fopen(filename, "r"); if (file == NULL) { printf("Error opening file.\n"); return; } int opcode, address; int i = 0; while (fscanf(file, "%d %d", &opcode, &address) == 2 && i < IM_SIZE) {…
Hello I need this little section of c++ code translated to MIPS assembly language MARS 4.2 Register usage:#################$a0: short-lived holder (in no-syscall areas)$a1: endPtr$a2: endPtr2$a3: still yet another short-lived holder$t0: oneInt$t1: used1$t2: used2$t3: used3$t4: hopPtr$t5: hopPtr2$t6: hopPtr3$t7: hopPtr22 or i or reply (non-overlappingly)$t8: numDup or probeIndex (non-overlappingly)$t9: numInts$v0: another short-lived holder (in no-syscall areas)$v1: yet another short-lived holder The code is :*(a1 + probeIndex) = oneInt;++used1;++i;
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