passing parameters to javascript function after function invocation?

Question

I am making a function that returns an arbitrary term of a linear sequence. The test cases look like this:

Test.assertEquals(getFunction([0, 1, 2, 3, 4])(5), 5, "Nope! Try again.");
Test.assertEquals(getFunction([0, 3, 6, 9, 12])(10), 30, "Nope! Try again.");
Test.assertEquals(getFunction([1, 4, 7, 10, 13])(20), 61, "Nope! Try again.");

I don't understand the function invocation. I have written this code to determine the function of the linear sequence and compute an arbitrary term, but I don't know how to pass my function the term to compute:

function getFunction(sequence) {
  var diff = sequence[1] - sequence[0];
  var init = sequence[0];

  return diff*arguments[1]+init;
}

arguments[1] doesn't access the term passed in after the parameters. How can I access the term (5) in the first example?

Test.assertEquals(getFunction([0, 1, 2, 3, 4])(5), 5, "Nope! Try again.");

you are only passing in your array to getFunction, there are no other arguments to get, hence why arguments[1] doesnt contain anything — Patrick Evans
– Patrick Evans, Commented Dec 19, 2014 at 21:53
As I read it, getFunction([...]) returns a function which takes 5 in argument (for the first test case), you won't reach the argument 5 from getFunction — topheman
– topheman, Commented Dec 19, 2014 at 21:55

dmamills · Accepted Answer · 2014-12-19 21:53:54Z

2

You need to return a function from getFunction() in order to chain your function calls like your tests are expecting

Something like this:

function getFunction(sequence) {
  var diff = sequence[1] - sequence[0];
  var init = sequence[0];
  return function(num) {
     return diff*num+init;
  }
}

answered Dec 19, 2014 at 21:53

dmamills

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1 Comment

user137717 Over a year ago

ahhh, so if I had an invocation like getSequence(stuff)(more stuff)(even more stuff); I would just have to keep nesting return statements that return functions?

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