recursion returning unexpected output

Question

I am starting with test(2,8)

I think the output should be 4 7 Instead I am getting 6 6 4 7 since p1 = p2 i.e. 6 = 6 the cout statement should not be performed. Why am I seeing the 6 6 ?

 using namespace std;

 void test(int p1, int p2);

  void  main()
  {
     test(2, 8);
     return ;
   }


  void  test(int p1, int p2)
  {
   if (p1 != p2)
   {
    p1 = p1 + 2;
    p2 = p2 - 1;
    test(p1, p2);
    cout << p1;
    cout << p2;
   }
  }

6 6 4 7 is correct , I guess you are not getting the recursive flow of the code — advocateofnone
– advocateofnone, Commented Mar 4, 2015 at 2:51
Your recursive call to test(p1, p2) is occurring before your cout << p1 and count << p2 that you think might be outputting 4 7. — lurker
– lurker, Commented Mar 4, 2015 at 2:52

advocateofnone · Accepted Answer · 2015-03-04 02:53:36Z

3

In first call to test p1 becomes 4 and p2 becomes 7. But before printing we again go into recursion , this time p1 becomes 6 and p2 also becomes 6. We again call recursion but as p1 is same as p2 it returns without printing anything ( does not enter if condition ). Then it prints 6 6 and when it returns to the most above level call to test function it prints 4 7. So output is 6 6 4 7.

answered Mar 4, 2015 at 2:53

advocateofnone

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3 Comments

Herbie Over a year ago

What is confusing me is the cout is in the IF statement, so since 6 = 6 how is the cout being executed?

M.M Over a year ago

@Herbie 4 7 is printed from within the if ( 2 != 8 ), and 6 6 is printed from within the if ( 4 != 7 )

advocateofnone Over a year ago

as @MattMcNabb said 6,6 is printed when arguments 4 , 7 are passed to test

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