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

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

No, they do not cut-short.

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

Only boolean operators && and || perform it.

This means that a short-cutting operator would have to be traditionally named &&= and ||=. Some languages provide them. C/C++ does not.

Answer 4

The code allTrue &= check_foo(); is equivalent to allTrue = allTrue & check_foo() In which you are using bitwise AND and no lazy evaluation is performed.

The bitwise AND must take two arguments who's binary representation has the same length, and useslogical AND operation to compare each corresponding pair of bits.

Answer 5

First: a &= b; is not the same as a = a && b;. a &= b; means a = a & b;. In C/C++ there is no a &&= b;.

Logical AND a && b is bit like a test for 1 bit. If the first "bit" is already 0, than the result will always be 0 no matter the second. So it is not necessary to evaluate b if the result is already clear from a. The C/C++ standard allows this optimization.

Bitwise AND a & b performs this test for all bits of a and b. So b needs to be evaluated if at least one bit in a would be non-zero. You could perhaps wish that if a==0, than b would not be evaluated, but this optimization is not allowed in C/C++.

Answer 6

Since & is a bit operation, check_foo()` will be evaluated irrespective of value of result

result = false;
result &= check_foo(); // check_foo() is needless called

However, check_foo() will not be called if you use && and result is false as in:

result = false;
result = result && check_foo(); // check_foo() is not called, the bitwise operator shortcircuits