function f2($n, $a)
{
if ($n == 0)
return 0;
return $a + f2($n-1, $a); // Changed return a... to return $a...
}
echo f2(3,4);
this outputs 12, I just don't understand why. Here's what I think, though:
obviously, $n and $a are the variables that will be take the place of the values. so like n is 3 and a is 4 in this example. the base case is the if statement, and it acts like a loop between the n. so since n is 3, and n is set to = 0 , then the code loops 3 times.
now i dont get the return a + f2(n-1, a) part. obviously that's where the math is calculated, but how?
it loops like this i think:
4 + f2(2, 4)
4 + f2(1, 4)
4 + f2(0, 4)
but what does it add to make the sum 12??
so what im asking is that if i understand it correctly and if not, what's actually happening and how is the output 12.
first call is f2(3,4)
f2(3,4)
|
return 4 + f(2,4)
|
return 4 + f(1,4)
|
return 4 + f(0, 4)
|
return 0; // and now unwinding the stack, 0 gets passed to the prev. return called.
|
return 4 + 0 // 4 gets passed to the prev. return called
|
return 4 + 4 + 0 // 8 gets passed to the prev. return called.
|
return 4 + 4 + 4 + 0 // f2(3,4) now returns a value of 12
looking at the call stack above , the first return called will be :
return 4 + 4 + 4 + 0
return a + call f2 , the + operator chains the a with the next function call, which at the end accumulates all your intermediate values in the first return called.