Chaining-friendly assertion that all elements are equal #4302

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()).