insertAll 方法

void insertAll(
  1. int index,
  2. Iterable<E> iterable
)
override

在此列表的 index 位置插入 iterable 中的所有对象。

这会增加列表的长度，并将所有后续对象移向列表末尾。

列表必须是可增长的。 index 值必须是非负数且不大于 length。

final numbers = <int>[1, 2, 3, 4];
final insertItems = [10, 11];
numbers.insertAll(2, insertItems);
print(numbers); // [1, 2, 10, 11, 3, 4]

实现

void insertAll(int index, Iterable<E> iterable) {
  RangeError.checkValueInInterval(index, 0, length, "index");
  if (index == length) {
    addAll(iterable);
    return;
  }
  if (iterable is! EfficientLengthIterable || identical(iterable, this)) {
    iterable = iterable.toList();
  }
  int insertionLength = iterable.length;
  if (insertionLength == 0) {
    return;
  }
  // There might be errors after the length change, in which case the list
  // will end up being modified but the operation not complete. Unless we
  // always go through a "toList" we can't really avoid that.
  int oldLength = length;
  for (int i = oldLength - insertionLength; i < oldLength; ++i) {
    add(this[i > 0 ? i : 0]);
  }
  if (iterable.length != insertionLength) {
    // If the iterable's length is linked to this list's length somehow,
    // we can't insert one in the other.
    this.length -= insertionLength;
    throw ConcurrentModificationError(iterable);
  }
  int oldCopyStart = index + insertionLength;
  if (oldCopyStart < oldLength) {
    setRange(oldCopyStart, oldLength, this, index);
  }
  setAll(index, iterable);
}