List Operations

View listOps.chpl on GitHub

The Chapel List module provides the implementation of the list type. Lists are useful for building up and iterating over a collection of values in a structured manner.

private use List;

config const quiet: bool = false;

We’ll start by declaring a list of int(64) and initializing it with the values contained in the range 1..8.

var lst1: list(int) = 1..8;

writeln("List 1 after init: ", lst1);

The most common operation performed on a list is list.pushBack(). The following code pushes some integers onto the end of our list.


The list type guarantees that appending a value to the end of a list will not invalidate references to existing values.

This is particularly useful in a parallel context. One task may push values to the end of a list while others index over existing values and use them in useful work.

for i in 1..8 do

writeln("List 1 after appends: ", lst1);

If a list is intended to be used in parallel, then the parSafe field must be set to true at initialization.

Let’s create a new list and list.pushBack() values to it in parallel.

var lst2: list(int, parSafe=true);

coforall tid in 0..3 with (ref lst2) {
  for i in 1..8 {
    const elem = tid * 8 + i;

Tasks spawned in a coforall loop aren’t guaranteed to execute in a fixed order. The contents of lst2 might be out of order even though our loop size is small (only 4 tasks).

We can call sort() on our list to be on the safe side.

if !quiet then
  writeln("List 2 before sort: ", lst2);


writeln("List 2 sorted: ", lst2);

We can create another new list with values that are copied from lst2.

var lst3 = lst2;

Before zippering lst2 and lst3 together, it would be good to vary their contents a little bit. Let’s list.popBack() the values from the first half of lst2 and list.pushBack() them to lst3.


The list.getAndRemove(idx) operation is O(n) worst case. If you remove a value from the front of a list, all the other values after it have to be shifted one to the left.

var count = 0;

while count < 16 {
  const elem = lst2.getAndRemove(0);
  count += 1;

writeln("List 2 after pops: ", lst2);
writeln("List 3 after appends: ", lst3);

Let’s ensure lst2 and lst3 have unique values. The list.remove() method takes a secondary argument called count which specifies how many instances of a value to remove. Passing 0 to count will remove every value matching the input from a list.

for elem in lst2 do
  lst3.remove(elem, 0);

writeln("List 3 after removes: ", lst3);

Even though lst2 and lst3 have no values in common, lst3 still has some duplicate values that could be removed. We can do that with a combination of list.remove() and list.count().


You should be careful not to modify the structure of a list while it is being iterated over, as this can cause the iterator to exhibit undefined behavior.

In this example, you’ll notice that we break out of our loop and start over if we happen to remove duplicate values during the current iteration.

var uniqued = false;

while !uniqued do
  for elem in lst3 {
    const count = lst3.count(elem);
    if count > 1 {
      lst3.remove(elem, count - 1);
    uniqued = true;

writeln("List 3 after removing duplicates: ", lst3);

Great. Now we can zipper our two lists together. Let’s double check our work and make sure that our two lists really share no values in common.


List iterators are not thread safe. They implicitly assume that their list is not being modified while iteration is occurring, and it is the user’s responsibility to abide by this assumption.

It is safe to modify the values of a list (i.e, references returned via indexing into a list) in a forall loop, but not the list itself.

forall (x, y) in zip(lst2, lst3) {

It seems like lst1 is just wasting memory at this point. Let’s go ahead and clear it using list.clear(). This will remove every value from the list and set its size to 0.


writeln("List 1 after clear: ", lst1);

We can use the list.pushBack() method to merge the contents of our lists together. Since lst1 is now empty, we can reuse it to save space.


writeln("List 1 after appends: ", lst1);

You’ll notice that the contents of lst1 are backwards. We could call list.sort() to fix this problem…or we can fix the contents of the list ourselves!


Indexing a list with a value that is out of bounds will cause the currently running program to halt. Be careful!

for i in 0..#(lst1.size / 2) {
  ref a = lst1[i];
  ref b = lst1[i + lst1.size / 2];
  const tmp = a;
  a = b;
  b = tmp;

writeln("List 1 after correction: ", lst1);

If you need to get the specific index of a value contained in a list, you can use the list.find() operator.


The list.indexOf() operator will halt if the search range specified by the arguments start and end falls outside the bounds of the list.

for x in lst2 {
  const idx = lst1.find(x);
  assert(x == idx+1);

for x in lst3 {
  const idx = lst1.find(x);
  assert(x == idx+1);

And finally, you can use the list.insert() method to insert a value at any position in a list.

As a trivial example, let’s insert the value -100 at the front of lst1.


Similar to list.getAndRemove(idx), the list.insert() operation is O(n) in the worst case.


The list.insert() method will halt if the index specified by the argument idx falls outside the bounds of the list.

lst1.insert(0, -100);

writeln("List 1 after inserting -100: ", lst1);