I am trying to wrap my head around recursive functions. I've tried a number of entry level exercises with no success. For example, I put this code into JS fiddle. I intended for it to take the sum of all numbers from 1 to n.

function sumAll(n) {
  if (n == 1 ) {
    return 1;
  } else if (n > 1) {
    return sumAll(n--)   n;
  }
}

console.log(sumAll(3));

feeding 3 to the function should give me '6'. I get an error, though.

CodePudding user response：

The -- suffix operator will evaluate to the original value of n, and the subtraction from n will happen afterwards (which is also not desired as you still want to do n). That means the recursive call gets the same value for n as the caller, and so you don't get closer to the end...

Don't use -- here, but -1:

function sumAll(n) {
  if (n == 1 ) {
    return 1;
  }
  else if (n > 1) {
    return sumAll(n-1)   n;
  }
}

console.log(sumAll(3));

CodePudding user response：

Perhaps you will enjoy a repeatable technique that can guide you through designing your recursive function. Let's solve sumAll using inductive reasoning -

1. If n is zero, return the empty sum, zero

2. (inductive) n is negative or positive. If n is negative, return the negative result of the subproblem sumAll(-n)

3. (inductive) n is positive. Return n plus the result of the subproblem sumAll(n - 1)

function sumAll(n) {
  if (n == 0)                  // 1
    return 0
  else if (n < 0)              // 2
    return -sumAll(-n)
  else
    return n   sumAll(n - 1)   // 3
}

console.log(sumAll(-10), sumAll(0), sumAll(10))
// -55 0 55

Here is sumAll written using a switch instead. It behaves identically but maybe you will find the syntax nicer to read -

function sumAll(n) {
  switch (true) {
    case n == 0:  return 0                  // 1
    case n < 0:   return -1 * sumAll(-n)    // 2 
    default:      return n   sumAll(n - 1)  // 3
  } 
}

console.log(sumAll(-10), sumAll(0), sumAll(10))
// -55 0 55

Here is sumAll again as a pure, functional expression -

const sumAll = (n = 0) =>
  n == 0                // 1
    ? 0
: n < 0                 // 2
    ? -1 * sumAll(-n)
: n   sumAll(n - 1)     // 3


console.log(sumAll(-10), sumAll(0), sumAll(10))
// -55 0 55