Reason for the existence of non-short-circuit logical operators

Question

When used with boolean operands, & and | become logical operators per Section 15.22.2 of the JLS. Unlike &&

Answer 1

Updated answer:

Apologies, I missed the word "logical" in your question even though it is there. (I've taken the liberty of emphasizing it a bit with an edit.)

Consider the case where you want any side-effects to always occur, regardless of whether the left-hand expression evaluates true or false. E.g., contrast:

if (foo() & bar()) {
    // Only call this if both operations returned true
}

with

if (foo() && bar()) {
    // Only call this if both operations returned true
}

Let's assume both foo and bar have effects that we want to have happen regardless of whether foo returns true or false. In the first one above, I know that bar will always get called and have its effect. In the latter, of course, bar may or may not get called. If we didn't have the non-short-circuit version, we'd have to use temporary variables:

boolean fooResult, barResult;
fooResult = foo();
barResult = bar();
if (fooResult && barResult) {
    // ...
}

You might argue (I probably would) that you should do that anyway, because it's way too easy to misread if (foo() & bar()), but there we go, a pragmatic reason for having non-short-circuit versions.

Original answer:

How would you propose & (or |) be a short-circuited operator? With && and ||, it makes sense because you're dealing with boolean conditions: They can be true or false, there are no shades of grey. But & and | deal with bits, not booleans. The result is a number. I mean, I guess & could not evaluate the right-hand side if the left-hand side were 0, and similarly | could not evaluate it if the left-hand side were all-bits-on for whatever the type was, but I don't see much point to making the one edge case of each operator significant (as compared to the 254 or more other cases).