19.Consider the following program. How many clock cycles required to executethe program in the pipelined implementation using stalling? (Concern the controlhazard only.) *[Program)i rmov $0,%raxi rmov $0,%rcx13. LOOP: стр %гсх,%гахje EXITinc %raxjmp LOOPi7. EXIT: hlti1.i2.14.i5.16.9101112

In general to execute an instruction it takes times and the system run with certain speed this speed…

Answered: (Program) i rmov $0,%rax i rmov $0,%rcx…

(Program) i rmov $0,%rax i rmov $0,%rcx i1. 12. 13. LOOP: стр %гсх, Хгах 14. i5. 16. i7. EXIT: hlt je EXIT inc %rax jmp LOOP

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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What the code below does? Consider the following code used to implement a new instruction: foo $t3,$t1,$t2: mask : . word 0xFFFFF83Fstart : la $t0 , masklw $t0 , 0 ( $ t 0 )la $t3 , s h f t rlw $t3 , 0 ( $ t 3 )and $t3 , $t3 , $ t 0andi $t2 , $t2 , 0 x 0 0 1 fsll $t2 , $t2 , 6or $t3 , $t3 , $ t 2la $t5 , s h f t rsw $t3 , 0 ( $ t 5 ) shftr: sll $t3 , $t1 , 0 What does foo does?
In <sys/param.h>, MAXPATHLEN is defined to be 1024 bytes.The for()loop here correctly bounds variable i to < 1023, such that when the loop hasended, no byte past npath[1023] may be written with \0. However, since i is alsoincremented in the nested statements here as ++i, it can become equal to 1024,and npath[1024] is past the end of the allocated buffer space.Then a null byte iswritten into npath[1024], overwriting the least significant byte of EBP.This canbe exploited as a off-by-one overflow.The bug was fixed by Which changing the logic?
Consider Context Switch time of 2 secs and modify below program accordingly Program: #include<stdio.h>#include<conio.h>using namespace std; int main(){int p[20],bt[20], su[20], wt[20],tat[20],i, k, n, temp; float wtavg, tatavg;printf("Enter the number of processes---"); scanf("%d",&n);for(i=0;i<n;i++){p[i] = i;printf("Enter the Burst Time of Process %d---", i); scanf("%d",&bt[i]);printf("System/User Process (0/1) ?---"); scanf("%d", &su[i]);}for(i=0;i<n;i++)for(k=i+1;k<n;k++)if(su[i] > su[k]){temp=p[i];p[i]=p[k];p[k]=temp;temp=bt[i];bt[i]=bt[k];bt[k]=temp;temp=su[i];su[i]=su[k];su[k]=temp;}wtavg = wt[0] = 0; tatavg = tat[0] = bt[0];for(i=1;i<n;i++){wt[i] =wt[i-1]+bt[i-1];tat[i] =tat[i-1]+ bt[i];wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\nPROCESS\t SYSTEM/USER PROCESS \tBURST TIME\tWAITING TIME\tTURNAROUND TIME"); for(i=0;i<n;i++)printf("\n%d \t\t %d \t\t %d \t\t %d \t\t %d ",p[i],su[i],bt[i],wt[i],tat[i]);printf("\nAverage…
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Solve the 8085 Write a program to load twenty memory locations starting from 8005H, where each location's content should increases by 2 over the previous one, however, the first location should contain 04H, assuming that the programs start at memory location 9009H?
A fragment of code is given to compare text in the field alpha with a text in the field beta: .data alpha db 12 dup(?) beta db ’abrakadabra’, 0 ; ::: .code ; ::: mov esi, 0 mov ecx, 12 ns: mov dl, alpha[esi] cmp dl, beta[esi] jne wrong ;=====MISSING==== loop ns ; ::: wrong: Choose missing instruction to insert in the code line marked by the comment “MISSING”.
Please help with the following in regards to Nand2Tetris, and hack code, so hack assembly and hack vm. There can be more that one answer to a question if so please explain why. 1d. Given a function called foo() that calls another external function bar() which in turn calls a second function called additup(). Indicate the VM line of code indicating the location in the program that control should be return to: a.@Foo.$bar. b. @Foo$bar$additup.ret.1 c. @Foo.bar.additup.ret.1 d. @Foo$bar$additup&return e. @Foo return,bar return.additup 1e. The return value of a function when it is completed is stored, in: a. calling Frame Arg[?] b. calling Frame LCL c. calling Frame SP d. working Frame Arg[?] e. working Frame LCL f. working Frame SP g. none of the above 1f. Every boolean expression of N inputs and one output can be implemented Circle the letter for ALL that apply. a. using a combination of D flip-flops, multiplexors and registers. b. by connecting the output directly to all…
Computer Science Extend your shell with pipelining (mandatory for teams of 3 or 4) The command $ cmd1 | cmd2 | cmd3 connects the standard output of cmd1 to the standard input of cmd2, and again connects the standard output of cmd2 to the standard input of cmd3 using the pipeline operator '|'. You will need to use the pipe() system call. Your shell should be able to handle up to three commands chained together with the pipeline operator (i.e., your shell should support up to two pipes pipelined together). This portion of the project should only require implementing support for a pipeline of 2 pipes/3 commands (no working with redirection). Your shell does not need to handle built-in commands implemented above (i.e., cd, exit, path, and myhistory) in pipeline. I need the code for this please.
This problem is adapted from an earlier edition of P&H, and should be submitted.Consider the following code used to implement a new instruction: foo $t3,$t1,$t2:mask : . word 0xFFFFF83Fs t a r t : l a $t0 , masklw $t0 , 0 ( $ t 0 )l a $t3 , s h f t rlw $t3 , 0 ( $ t 3 )and $t3 , $t3 , $ t 0a ndi $t2 , $t2 , 0 x 0 0 1 fs l l $t2 , $t2 , 6o r $t3 , $t3 , $ t 2l a $t5 , s h f t rsw $t3 , 0 ( $ t 5 )s h f t r : s l l $t3 , $t1 , 0Add meaningful comments to the code and write a brief (2 sentence max) description of what foo does. Thisis not the same as saying how it does it - e.g., when asked to describe what a pedestrian is doing, you wouldsay they are walking, not that they are ilfting their left leg, angling it forward, putting it down, . . ..State at least one reason why writing “self-modifying code” such as this is a bad idea (and often times notactually allowed by the operating system)?
The following code attempts to find all the prime numbers between 2 and n. 1. When compiled, why is the following error produced?: Undefined symbols for architecture x86_64: "_main", referenced from: implicit entry/start for main executableld: symbol(s) not found for architecture x86_64 2. Fix code so that it compiles with no errors #include <stdio.h> #include <stdlib.h> #include <math.h> #include <sys/time.h> //helper function: computes wall clock time double getTime(struct timeval ts, struct timeval te){ doubletime = te.tv_sec - ts.tv_sec + (te.tv_usec - ts.tv_usec)/1.e6; returntime; } //helper function: allocates an array of a specified length and returns a pointer int * allocateArray(int len) { int * result = malloc(len * sizeof(int)); returnresult; } //helper function: prints out elements of array separated by spaces void printArray(int * arr, int len) { inti; for (i = 0; i < len; i++) { printf("%d ", arr[i]); } printf("\n"); } //helper…
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