JavaScript If index is out of bounds of array, set index to closest boundary index array bounds matlab,index outside bounds array c#,index

Let's say we have an array that can have a length of 1 to ??

We have a way of navigating this array up or down by clicking a button that either navigates upward or downward, which is displayed in the DOM (think pagination).

We would want our navigation to stop when either we are out of the lower bound, or stop when our navigation is out of the upper bound.

When we reach the lower-bound we would know this by the value -1

When we reach the upper-bound we would know this by the value ( > array.length - 1 ).

Our function accepts the parameter index as shown

function goToChapter( index ){
  // compute value of index to within bounds.
  return array[ index ];
}

Let's say all we want to do is use the computed within bounds index to get a value from the returned array.

Examples

var array ["chapZero", "chapOne", "chapTwo", "chapThree"];

goToChapter( 2 );
=> "chapTwo;

goToChapter( 4 );
=> "chapThree"; // because four is out of upper bound so index became three.

goToChapter( -1 );
=> "chapZero"; // because -1 is our of lower bound so index became zero.

I realize this can be done with a few if statements, I am looking for a way to use some sort of math formula that perhaps

  1. converts any negative number to 0;
  2. leaves any positive number alone;
  3. uses the length of the array to determine when to return the upper limit.

UPDATE

I'm going with Andand's answer. I've added enough to copy paste the solution into the console if you find this helpful and want to test it out.

var array = [0,1,2,3,4];

function goToChapter( index, array ){
  return array[(Math.min( array.length - 1, Math.max( 0, index )))];
}

goToChapter( -1, array ); => 0;

goToChapter( 3, array ); => 3;

goToChapter( 8, array ); => 4;

Thanks Andand.

NOTE:

I also found this helpful, but not as fast.

function setWithinArrayBounds( index, array ){
  return !!index ? 0 : index > ( array.length - 1 ) ? ( array.length - 1 ) : index;
}
Answer:1

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