description

# lastIndexOf method Null safety

int lastIndexOf (

1. Object? element,
2. [int? start]

)
override

Returns the last index of `element` in this list.

Searches the list backwards from index `start` to 0.

The first time an object `o` is encountered so that `o == element`, the index of `o` is returned.

    List<String> notes = ['do', 're', 'mi', 're'];
    notes.lastIndexOf('re', 2); // 1

If `start` is not provided, this method searches from the end of the list.

    notes.lastIndexOf('re');  // 3

Returns -1 if `element` is not found.

    notes.lastIndexOf('fa');  // -1

## Implementation

    int lastIndexOf(Object? element, [int? start]) {
      if (start == null || start >= this.length) start = this.length - 1;

      // TODO(38493): The previous line should promote.
      if (start == null) throw "!";

      for (int i = start; i >= 0; i--) {
        if (this[i] == element) return i;
      }
      return -1;
    }