Do the &= and |= operators for bool short-circuit?

When writing code like this in C++:

bool allTrue = true;
allTrue = allTrue && check_foo();
allTrue = allTrue && check_bar();

check_bar() will not be evaluated if check_foo() returned false. This is called short-circuiting or short-circuit evaluation and is part of the lazy evaluation principle.

Does this work with the compound assignment operator &=?

bool allTrue = true;
allTrue &= check_foo();
allTrue &= check_bar(); //what now?

For logical OR replace all & with | and true with false.

From C++11 5.17 Assignment and compound assignment operators:

The behavior of an expression of the form E1 op = E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once.

However, you're mixing up logical AND which does short-circuit, and the bitwise AND which never does.

The text snippet &&=, which would be how you would do what you're asking about, is nowhere to be found in the standard. The reason for that is that it doesn't actually exist: there is no logical-and-assignment operator.

The short-circuit (i.e. lazy) evaluation is only for logical && and ||. Bitwise & and | evaluate both arguments.

No, they do not cut-short.

Note that the &= and |= operators are formed as &+= and |+=. Bit operators & and | does not perform shortcut evaluation.

Only boolean operators && and || perform it.

It means, that a shortcutting operator would have to be traditionally named &&= and ||=. Some languages provide them. C/C++ does not.