Function removeFirstWhere

removeFirstWhere<T>(array, fn, start?, stop?): {
index: number;
value: T | undefined;
}
Remove the first occurrence of a value which matches a predicate.
Type Parameters
- T
Parameters
- array: T[]
  The array of interest.
- fn: ((value, index) => boolean)
  The predicate function to apply to the values.
  - - (value, index): boolean
    - Parameters
      value: T
      index: number
      Returns boolean
- start: number = 0
  The index of the first element in the range to be searched, inclusive. The default value is 0. Negative values are taken as an offset from the end of the array.
- stop: number = -1
  The index of the last element in the range to be searched, inclusive. The default value is -1. Negative values are taken as an offset from the end of the array.
Returns {
index: number;
value: T | undefined;
}
The removed { index, value }, which will be -1 and undefined if the value is not contained in the array.

Notes
If stop < start the search will wrap at the end of the array.

Complexity
Linear.

Example
```
import { ArrayExt } from '@lumino/algorithm';

function isEven(value: number): boolean {
  return value % 2 === 0;
}

let data = [0, 12, 23, 39, 14, 12, 75];
ArrayExt.removeFirstWhere(data, isEven);     // { index: 0, value: 0 }
ArrayExt.removeFirstWhere(data, isEven, 2);  // { index: 3, value: 14 }
ArrayExt.removeFirstWhere(data, isEven, 4);  // { index: -1, value: undefined }
```
- index: number
- value: T | undefined
- Defined in array.ts:1348