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Chaining-friendly assertion that all elements are equal #4302
Description
Feature summary
Provide a chaining-friendly way to assert that all elements are the same:
assertThat(list).allElementsTheSame();The current best-practice requires using allMatch() and referencing the first element of the list:
assertThat(list).allMatch(e -> e.equals(list.get(0));. However that patter falls short in more complex use cases where the list is produced by filteredOn() and extracting() sequences and cannot be easily referenced.
Example
My current real-world test that verifies that the output table is aligned:
List<String> dataLines = Arrays.stream(output.split("\n"))
.filter(s -> s.contains("|"))
.toList();
assertThat(dataLines)
.hasSize(20)
.allMatch(s -> s.matches(".* \\| .* --- \\w+"));
List<int[]> pipePositions = dataLines.stream()
.map(s -> new int[] {s.indexOf('|'), s.indexOf("---")})
.toList();
assertThat(pipePositions)
.allMatch(p -> Arrays.equals(p, pipePositions.get(0)));Note how I cannot join those 2 checks into one because i have to reference the first element.
How it could look like:
assertThat(output.split("\n"))
.filteredOn(s -> s.contains("|"))
.hasSize(20)
.allMatch(s -> s.matches(".* \\| .* --- \\w+"))
.extracting(s -> new int[] {s.indexOf('|'), s.indexOf("---")})
.allElementsTheSame()Special Consideration
As shown in my example, sometimes equals() check on elements is not adequate and a special check is needed (for example Arrays.equals()).
Activity
joel-costigliola commentedon Jun 27, 2026
We can add that as I don't see another way of doing such assertions.
As a side note, instead of
you can use
@voddan would you been keen to contribute this ?
voddan commentedon Jun 27, 2026
@joel-costigliola yes, I would love to contribute this!
I can suggest several types of API:
Which method signatures and method names do you think are best?
joel-costigliola commentedon Jun 27, 2026
Let's start with
allElementsAreEqual(not the same, which would mean checking by references) and discuss the pair assertion in another issue.scordio commentedon Sep 12, 2026
To stay closer to the existing
containsOnlyNulls, what aboutcontainsOnlySameElementcontainsOnlyEqualElements?