Recursive function with for

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Recursive function with for

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#1 by (2 votes)
#2 by (1 votes)
#3 by (-2 votes)

2

I'm studying recursive functions and I had a question when I was looking at some algorithms that use recursive function and for . I decided to create a simple example and print some values on the screen to try to understand how the flow of the program is and I ended up getting more involved.

Could anyone explain what happens?

Sorry if I'm not clear enough, I'm a beginner and I can not express myself very well.

Follow the code:

function x(n){
    console.log('before , n: ',n);
    if(n === 0)return;
    for(var i = 1; i < 3; i++){  
        console.log('for i:',i);  
        x(n-1);
    }
    console.log('after, n: ',n);
}
x(3);

Follow print:

javascript

asked by anonymous 31.01.2017 / 18:12

3 answers

1

Your question is "Why is for saving the value of i = 2 after the base case is reached?"

This is because javascript has function scope instead of block scope , with initialization of i being done within the function where for is, so it knows the value.

31.01.2017 / 19:25

-2

What he is doing is: for every n , he executes this cycle 3 times. Try changing the cycle for to for (var = i ; i < n ; i++)

31.01.2017 / 18:34

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score 2 · Accepted Answer

We call recursion or recursion when a function calls itself.

In a recursive function, each call is created in memory a new occurrence function with commands and variables "isolated" from previous occurrences, so a new scope for each function call. / p>

When the line x(n-1) is executed, the current instance waits for a response of the new instance, ie, you must wait for the result of the x(n-1) operation for the next operation to be performed.

In practice the following operations occur:

Call the function x(n) .

Prints the value of n (before) in the console.

Ends occurrence if n equals 0 by returning a response to the previous instance (if any).

Cycle for .

Prints the value of i in the console.

Starts a new instance x(n-1) (returns to point 2) and waits for the response.

Returns to point 4 if i < 3 .

Prints the value of n (after) in the console.

Notice that at point 6 we create a new occurrence (n-1) and this causes the current occurrence (n) to wait for the response of the new occurrence before passing to the point 7 .

A step by step example for n=2 according to the scheme above:

1. x(2);
2. print "before, n: 2"
3. N > 0
4. For i=1
5. print "for i: 1"
6. x(1) //n-1 em que n=2
    1. Iniciou uma nova ocorrência
    2. print "before, n: 1"
    3. N > 0
    4. For i=1
    5. print "for i: 1"
    6. x(0) //n-1 em que n=1
        1. Iniciou uma nova ocorrência
        2. print "before, n: 0"
        3. N <= 0
        8. print "after, n: 0"
        [Termina a ocorrência]
    4. For i=2
    5. print "for i: 2"
    6. x(0) //n-1 em que n=1
        1. Iniciou uma nova ocorrência
        2. print "before, n: 0"
        3. N <= 0
        8. print "after, n: 0"
        [Termina a ocorrência]
    4. For i=3
    8. print "after, n: 1"
    [Termina a ocorrência]
4. For i=2
5. print "for i: 2"
6. x(1) //n-1 em que n=2
    1. Iniciou uma nova ocorrência
    2. print "before, n: 1"
    3. N > 0
    4. For i=1
    5. print "for i: 1"
    6. x(0) //n-1 em que n=1
        1. Iniciou uma nova ocorrência
        2. print "before, n: 0"
        3. N <= 0
        8. print "after, n: 0"
        [Termina a ocorrência]
    4. For i=2
    5. print "for i: 2"
    6. x(0) //n-1 em que n=1
        1. Iniciou uma nova ocorrência
        2. print "before, n: 0"
        3. N <= 0
        8. print "after, n: 0"
        [Termina a ocorrência]
    4. For i=3
    8. print "after, n: 1"
    [Termina a ocorrência]
4. For i=3
8. print "after, n: 2"
[Termina a ocorrência]

In the above list, I used spaces horizontally to simulate a new occurrence of the x function.

If we only extract the lines in which "print" appears, we will have the following:

before, n: 2
for i: 1
   before, n: 1 
   for i: 1
      before, n: 0
      after, n: 0
   for i: 2
      before, n: 0
      after, n: 0
   after, n: 1
for i: 2
   before, n: 1
   for i: 1
      before, n: 0
      after, n: 0
   for i: 2
      before, n: 0
      after, n: 0
    after, n: 1
after, n: 2

Although the first occurrence is x(2) , after, n: 2 only appears at the bottom of the list because new occurrences appeared in the middle and fired console.log before returning a response.