Question

int j(int i) {
    if (i==0) return 2;
    return i*j(i-1);
}

calling j(4) gives 48.

I feel like I do understand the code, but don't know how 48 was achieved.

This is how I interpret it,

int j(4){

4 is not equal to 0 so it won't return 2

4*j(i-1) = 4*j(3)

then i go back to the top and repeat

3*j(2)
2*j(1)
1*j(0)

now i return 2.

This is where am lost, i'm not quite sure what to do after this step and how the answer to j(4) is 48.

Était-ce utile?

La solution

The evaluation can be understood like this

j(4)
4 * j(3)
4 * (3 * j(2))
4 * (3 * (2 * j(1)))
4 * (3 * (2 * (1 * j(0))))   // When input to j is 0, returns 2
4 * (3 * (2 * (1 * 2)))
4 * (3 * (2 * 2))
4 * (3 * 4)
4 * 12
48

Till recursion reaches j(0), it goes deeper and deeper. But when it reaches it, it begins to unwind the recursion. And the most important thing is,

if (i==0) return 2;

is called the base condition of your recursion.

Autres conseils

j(0) clearly returns 2, right?

So j(1) returns 1*j(0), and j(0) is 2. So j(1) returns 2*1, or 2.

j(2) returns 2*j(1) or 2*2 or 4.

j(3) returns 3*j(2) or 3*4 or 12.

j(4) returns 4*j(3) or 4*12 or 48.

so this is calling

return 4 * 3 * 2 * 1 * 0 (no not 0 use 2)

result 48

You can trace through it:

j(4) = 4 * j(3) = 4 * (3 * j(2)) = 4 * 3 * 2 * j(1) 
     = 4 * 3 * 2 * 1 * j(0) = 4 * 3 * 2 * 1 * 2 = 48
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