Output of this code

Question

Output of this program -

 

int rec(int x)

{

    static int f;
    if(x == 1)
       return(1);
    else  __Y__  ;

    return f ;
}

 

What is the value returned by rec(5)

​a)   when  Y  is     f = f * 1 + rec(x-1) ;
b)    when  Y  is     f = f * x  + rec(x-1) ;

What is proper way to solve this ?

Comments

i was correct sir :(

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yes u r correct,  :)

still there is some need for reasoning, for particular behaviour.. as I have given 7 cases in above comment, and need to know why a statement is behaving in a particular manner..

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@Himanshu1 sir, I can't understand this line - 

f = f * x + rec(x-1) ;    gives  1  // not using updated value of f

Can you please explain a bit more? 

Answer 1

int rec(int x)

{

    static int f;
    if(x == 1)
       return(1);
    else 

  f = f * 1 + rec(x-1)  ;

     return f ;
}

 

in this 

  f = f * 1 + rec(x-1) ;

is equal to writing f=f+rec(x-1) so f has to be computed when rec(x-1) value is returned to function 

Since f is declared as static so it will be initialized to 0

 

so              rec(5)=8

                                 f=f+rec(4)=4+4

                                 f=f+rec(3)=2+2

                                 f=f+rec(2)=1+1

                                 f=f+rec(1)=0+1   return (rec(1))=1

 

2)

 

int rec(int x)

{

    static int f;
    if(x == 1)
       return(1);
    else 

  f = f * x + rec(x-1)  ;

     return f ;
}

 

in this 

  f = f * x + rec(x-1) ;

here multiplication will be done first and then function will be called ,so returning value of function will not change the result and thus changed value of f  will not reflect.

Since f is declared as static so it will be initialized to 0

 

so              rec(5)=1

                                 f=f*x+rec(4)=0+1

                                 f=f*x+rec(3)=0+1

                                 f=f*x+rec(2)=0+1

                                 f=f*x+rec(1)=0+1   return (rec(1))=1

 

 

 

Answer 2

My Answer is

Comments

k fine thank u

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@ shivanisrivarshini if you write 

int rec(int x)

{

    static int f;
    if(x == 1)
       return(1);
    else 

  f =  rec(x-1)+f * x   ;

     return f ;
}

then you will get 120 bcz then your procedure is correct bcz then value of f is dependent on returning value of rec so first rec will be called and thus latter portion will go to hold and will use the changed of f when it is popped from the stack.

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but f=f+rec(x-1) here also f is  not depending on rec()  then cant we have as follows

rec(5)=f+rec(4)=0+1=1

rec(4)=f+rec(3)=0+1=1

rec(2)=f+rec(1)=0+1=1

rec(1)=1

then why 8???