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In what order will the numbers 1-4 be logged to the console when the code below is executed? Why?

+4 votes
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In what order will the numbers 1-4 be logged to the console when the code below is executed? Why?
posted Apr 8, 2016 by Vrije Mani Upadhyay

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1 Answer

+1 vote
 
Best answer

The values will be logged in the following order:

1
4
3
2

Let’s first explain the parts of this that are presumably more obvious:

1 and 4 are displayed first since they are logged by simple calls to console.log() without any delay
2 is displayed after 3 because 2 is being logged after a delay of 1000 msecs (i.e., 1 second) whereas 3 is being logged after a delay of 0 msecs.
OK, fine. But if 3 is being logged after a delay of 0 msecs, doesn’t that mean that it is being logged right away? And, if so, shouldn’t it be logged before 4, since 4 is being logged by a later line of code?

The answer has to do with properly understanding JavaScript events and timing.

The browser has an event loop which checks the event queue and processes pending events. For example, if an event happens in the background (e.g., a script onload event) while the browser is busy (e.g., processing an onclick), the event gets appended to the queue. When the onclick handler is complete, the queue is checked and the event is then handled (e.g., the onload script is executed).

Similarly, setTimeout() also puts execution of its referenced function into the event queue if the browser is busy.

When a value of zero is passed as the second argument to setTimeout(), it attempts to execute the specified function “as soon as possible”. Specifically, execution of the function is placed on the event queue to occur on the next timer tick. Note, though, that this is not immediate; the function is not executed until the next tick. That’s why in the above example, the call to console.log(4) occurs before the call to console.log(3) (since the call to console.log(3) is invoked via setTimeout, so it is slightly delayed).

answer Jun 7, 2016 by Deepak Negi
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+3 votes

What will be the answer of below code?

var myObject = {
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+2 votes

I have function in python,(Assume that i have imported all necessary module),

 def DL_Iperf(args):
        ssh=paramiko.SSHClient()
        ssh.set_missing_host_key_policy(paramiko.AutoAddPolicy())
        ssh.connect(server_ip,username="root",password=Password)
some_code

This function is actually a thread and it will be created as many no of UE i have, (Ex: if i have 1 UE than 1 Thread will be created),

So, if i have 1 UE/ 2 UE than its working but if i have 3 UE then it is failing, with error "Paramiko : Error reading SSH protocol banner",

Below is the stderr of the script,

    No handlers could be found for logger "paramiko.transport"

    Unhandled exception in thread started by <function DL_Iperf at 0x02B8ACF0>
    Traceback (most recent call last):

    File "C:\Users\qxdm-5\Desktop\Chirag\LTE_11_Perfect_Working\TCP_Latest_2\Windo
    ws_UE\slave.py", line 379, in DL_Iperf

    ssh.connect(ServerIp,username="root",password=Pwd)

    File "build\bdist.win32\egg\paramiko\client.py", line 295, in connect

    File "build\bdist.win32\egg\paramiko\transport.py", line 451, in start_client

paramiko.SSHException: Error reading SSH protocol banner

From some reference i found that this is because of some network related issue, but my question is if it network related then why everytime in 3rd call of the function i am getting this error? And how do i resolve it?

+3 votes

1.

main()
{
printf("%x",-1<<4);
}

2.

main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
...