Feedback Nodes & Nested FOR Loops

[SercoSteveB]

What is Numeric Value Out following execution of the VI?

a) 0

b) 2

c) 4

d) 6

[crossrulz]

C

[sathishp91]

C

[nik35324]

C

[LordNobady]

C

[A.Bernau]

I get C too.
(9 minus 5 => 4)
I was not able to solve this just in my head.
I had to print it out and had to use pens with diffrent coulors to keep track of.
So I am not sure if I could solve this in an environment of a real exam from NI ...
I should try to solve more Daily CLADs from the past 😉

@crossrulz: OK, thank you. I try to remeber your words next time if the question is from similiar type.

[Mohammed_Atef]

C

[crossrulz]

Andreas23 wrote:

I was not able to solve this just in my head.

                   

You need to learn to simplify things then...

The inner loop runs 6 times (2*3=6) and the feedback node is initialized to 3.  So for the top, you get 2*3*1+3 = 9.  Since the bottom loop reinitializes the feedback node each time the outter loop is ran, you get 2*1*1+3 = 5.  9-5=4.

[sriveni]

C

[MrStevenUND]

C

[dellmore]

C. The top for loop has it's outer loop initialized to three and it won't re-initialize until the loop is called again else where. Since that doesn't happen the end result should be 9.

The bottom for-loop has it's inner loop initialized to three and every time it's called the inner loop re-initializes. This happens every time the bottom outer loop is calls it so the end result should be 5. Subtracting the two should result in 9-5=4.

-Ricky

[hemanth.wnp@gmail.com]

Ans: C

[mini09]

C

[SercoSteveB]

Answer: C.  Nice one crossrulz, sathishp91, nik35324, LordNobady, Andreas23, Mohammed_Atef, sriveni, MrStevenUND, dellmore, hemanth.wnp@gmail.com & gnshmrthy.

[s.vishal91]

steve can u explain the feedback nodes...I don't get it.

[crossrulz]

Feedback Nodes work just like shift registers: they pass data from one loop iteration to the next.

The point to notice here is where the shift register is being initialized.

The top diagram has the Feedback Node being initialized on the outer loop.  This would be equivalent to having a shift register on both loops and initializing the outer loop's shift register.

The botton diagram has the Feedback Node being initialized on the inner loop.  This would be equivaltent to having a shift register on just the inner loop.  This also means that the feedback node is re-initialized with each iteration of the outer loop.

Here is what the equivalent would be using shift registers:

[s.vishal91]

crossrulz: Thanx for the prompt and detailed explanation. Could you please explain the couple of questions on initialization of property nodes also.

[s.vishal91]

sorry feedback nodes

[hameedkdnl]

@CrossRulz, Thank You For your breif Explanation about FB Node...You are Awesome!!!!..

[crossrulz]

hameedkdnl wrote:

                       

@CrossRulz...You are Awesome!!!!..

                   

Wait, are you saying I'm not awesome?  That hurts...

[hameedkdnl]

@Cross Rulz..Just Kidding...

[SercoSteveB]

        .  That's priceless.  I will need to remember that one.

[ashwinilele]

C

[M.P.Bharghavi]

Ans:C

4