About

Pursuing M.Tech in Computer Science & Engineering at IIT Kharagpur and Editor at Techtud.

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Alma Mater:

B.tech in Compter Science and Engineering
Kamla Nehru Institute of Technology, Sultanpur
2012 to 2016

Experience:

Not updated.
Techtud Editor
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5 months 2 weeks ago
All the Best for GATE 2018 :)
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Techtud Editor
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5 months 3 weeks ago
Tomorrow is the last date for attempting VirtualGATE-2.
http://virtualgate.in/login/index.php
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Techtud Editor
Example text
6 months 1 week ago

c stands for Complement here.

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Techtud Editor
Example text
6 months 1 week ago

@vikash957 at x= -3, function is not defind ( devide by zero).

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Techtud Editor
Multichoice
6 months 1 week ago
@reenakandari
Question is about non-preemptive kernel design, where a running process can be preempted only when it is being forced for it.
Timer interrupts and Disk interrupts will wait untill the process is running. A blocking system call or an exit request are the high priority kernel requests, so these can not be hold on wait.
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Fine then, let's go line by line:

1.if (fork() == 0)

2. { a = a + 5; printf("%d,%d\n", a, &a); }

3. else { a = a –5; printf("%d, %d\n", a, &a); } 

Process P will execute line1. fork() will cause creation of a new child process say C1
Now Both P and C1 will start execution from the next instruction in line 1 which is the comparision of the returned value of fork() with zero.
For parent process P, this comparision is false as value returned by fork will be greater than 0(Process id of child process). So P will execute the else block.
For child process C1, Condition fork()==0 is true. So Cwill execute the if block.
Now u, v be the values printed by the parent process, and x, y be the values printed by the child process.
u= a-5, v=address of a (in parent process)
x= a+5, y=address of a (in child process)
Note one thing here: In virtual memory environment, all the processes start execution from the virtual address zero and hence both the process will have same logical address of the variable 'a' (physical address will be different).
Hence u + 10 = x and v = y is the correct option.

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@shweta1920 
Then run the processes in increasing order of run time.

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14 Jun 2017 - 1:41pm

https://www.d.umn.edu/~gshute/arch/register-renaming.xhtml
Please inform if you still have any doubt.

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The best scheduling algorithm can be Round Robin here.

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If cheksum will not match, the receiver will discard the packet and will not generate acknowledgment for that packet. So sender will send it again. There is no way, the receiver can say that checksum is corrupted.

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50 percent rule:
Consider a process in memory. About 50% of the time there are allocations and the rest are deallocations. So each process has a hole above it 50% of the time.
If there are n processes, then the expected number of holes is n/2.
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11 Jun 2017 - 4:37pm

No previous year paper is available on the Internet.

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