具有双输出的递归函数

Question

I am trying to make an recursive function. i will be an integer, however the output will be an double. I know the function cannot take an double to start, however when the function reaches it maximum "depth", i will still be an integer. When it will go "up" the output will become an double, but i do not understand why the function will not work.

long sum1(int i) {
  if(i==1) {
    return 1;
  } else {
    return sum1(i-1) + 1/i;
  }
}

This is my code to perform this serie:

1. series1(i) = 1 + 1/2 + 1/3 + . . . 1/i

The output does not give an error, but the result is not compute correctly.

Answer 1

First of all 1/i will always be 0 when i > 1, since you are dividing two int s.

Second of all, since the return type is long , any fractions will be truncated anyway.

You should return a double for an accurate result:

double sum1(int i) {
  if(i==1) {
    return 1.0;
  } else {
    return sum1(i-1) + 1.0/i;
  }
}

Answer 2

Well the problem for the wrong calculation is that sum1(i-1) + 1/i; results in an integer calculation. And 1/i when i is integer >1 is always = 0.

Just use double for the recursion and you can pass int argument.

double sum1(double i) {
          if(i==1) {
            return 1;
          } else {
            return sum1(i-1) + 1/i;
          }
        }