递归调用的返回值是多少

Question

java version "1.8.0_92"

I have the following code I am trying to understand and trace.

The part I don't understand is the * a at the end. When does that multiplication get called? And what is the return value of power1(a, n - 1);

Is it n - 1 * a

 public static double power1(double a, int n) {
        double result;
        if(n == 0) {
            return 1;
        }
        else {
            result = power1(a, n - 1) * a;

            return result;
        }
    }

Answer 1

You can modify the code to print a trace of the recursion. That can help you understand what is going on.

/**
 * Print string form of `o`, but indented with n spaces
 */
private static void printIndented(int n, Object o) {
    while (n-->0) System.out.print("  ");
    System.out.println(o);
}

/* 
 * Added a third param `d` to keep track of the depth of the recursion
 */
public static double power1(double a, int n, int d) {
    // Entering a "possible" recursive call
    printIndented(d, "call power1, a=" + a + ", n=" + n + ", d=" + d);

    double result;
    if(n == 0) {
        // Returning from the base case, this should have the largest depth.
        printIndented(d, "return 1.0");

        return 1;
    }
    else {
        result = power1(a, n - 1, d + 1);

        // Return from intermediate recursive calls, we print
        // the value of power1(a, n-1) as well.
        printIndented(d, "return " + result + " * " + a);
        return result * a;
    }
}

public static void main(String [] args) {
    System.out.println(power1(1.4, 3, 0));
}

Output

call power1, a=1.4, n=3, d=0
  call power1, a=1.4, n=2, d=1
    call power1, a=1.4, n=1, d=2
      call power1, a=1.4, n=0, d=3
      return 1.0
    return 1.0 * 1.4
  return 1.4 * 1.4
return 1.9599999999999997 * 1.4
2.7439999999999993

As you can see, the value from the inner return becomes the result in the outer return statement.

Answer 2

Return value of power1(a,n-1)

We can see that power1 is defined as public static _double_ power1(double a, int n) This means that on the line

result = power1(a, n - 1) * a;

the type will be :

double = double * double;

When the multiplication gets called

We start our function with n and a . a will be constant in this implementation. It's given straight as to the next recursive call.

Yet n varies.

We call power1(a,n) . It checks first if n == 0; It is not. So we go to the else part of the if. Time to calculate result .

To know what the value of result is, we need power1(a,n-1). We proceed. n-1 can be 0 or can not be.

If it's 0, we return 1, and we now know that power1(a,n-1) = 1. We can now multiply it with a.

If it's not 0, then we need a new result, seeing as we're in a completely seperate call of the power1 method. We now need power1(a,n-2). We check again for 0. If n-2 == 0, return 1 to get the caller to calculate power(a,n-1). If not go down another call, now with n-3... Etc

In terms of the timing of calling the multiplication.

It's gonna call all the (n) recursive calls first, before doing n multiplications.

Answer 3

If you call power1 with a = 2 and n = 3 , this is what will happen:

result = power1(2, 2) * 2;
power1(2, 2) = power1(2, 1) * 2;
power1(2, 1) = power1(2, 0) * 2;
power1(2, 0) = 1

In the above example, the * a happens three times in total. It is called whenever power1(a, n) returns a value. The first power1(a, n) to return a value will be power1(2, 0) (this is because n = 0 is your "base" case). Then power1(2, 1) will return 2. Then power1(2, 2) will return 4. Then your initial call to power1(2, 3) will return 8.

Answer 4

First you call power (a,n-1) until n == 0 which is your base case .

Then the 1 gets returned and is multiplied by the value of a. Then a is returned to the previous call where it is multiplied by a . The process is repeated N times and thus you get A^n. I would recommend you go throught one example of this code giving it some initials values and tracing the code using pen and paper.