Thread: Quick Code analysis please

Can someone please give me a quick code analysis of what is outputted when the method f1(5) is called?

 public in f1( int x )
 
 {
 
 if ( x == 0 )
 
 return 0;
 
 else
 
 return f2( x - 2 );
 
 }
 
 public in f2( int x )
 
 {
 
 if ( x == 1 )
 
 return 1;
 
 else 
 
 return f1( x + 1 ) + x;
 
}

What happens when the code is compiled and executed?
If you want to see the values of intermediate results, add println statements to print them.

postfix/prefix i think(the order you do the math), if your wanting it to add the x either put it in the brackets that exist or put another set of brackets around the +1. my thinking is that its just returning the value of x along with the result from the function call, as the function take only one argument, unless its in brackets it concatenating(adding) the results together; i.e when all is said and done, we will probably end up with 11, as in x = 1(result from all the function calls), x =1 put together....

5 enters f2 as 3, goes back to 4 and becomes 2 goes up to 3 and finally finishes at 1, essentially taking 2 originally and then going down one at a time.

No, it cannot be ignored.

f1(x+1) returns some number. Simply substitute that result, perform the +x, and return the result.

Try stepping through the code in debug mode or adding print statements to your code to see how the result as each line of code gets executed.