Using Bitwise Operators

As an example of using the bitwise operators let’s assume a teacher has 30 students in a class. Each week the class is given a pass/fail quiz. We’ll track the results of each quiz using one bit per student to represent the pass or fail grade on a given test. We might represent each quiz in an unsigned integral value:

unsigned long quiz1 = 0; // we'll use this value as a collection of bits

We define quiz1 as an unsigned long. Thus, quiz1 will have at least 32 bits on any machine. We explicitly initialize quiz1 to ensure that the bits start out with well-defined values.

The teacher must be able to set and test individual bits. For example, we’d like to be able to set the bit corresponding to student number 27 to indicate that this student passed the quiz. We can indicate that student number 27 passed by creating a value that has only bit 27 turned on. If we then bitwise OR that value with quiz1, all the bits except bit 27 will remain unchanged.

For the purpose of this example, we will count the bits of quiz1 by assigning 0 to the low-order bit, 1 to the next bit, and so on.

We can obtain a value indicating that student 27 passed by using the left-shift operator and an unsigned long integer literal 1 (§ 2.1.3, p. 38):

1UL << 27 // generate a value with only bit number 27 set

1UL has a 1 in the low-order bit and (at least) 31 zero bits. We specified unsigned long because ints are only guaranteed to have 16 bits, and we need at least 27. This expression shifts the 1 bit left 27 positions inserting 0 bits behind it.

Next we OR this value with quiz1. Because we want to update the value of quiz1, we use a compound assignment (§ 4.4, p. 147):

quiz1 |= 1UL << 27; // indicate student number 27 passed

The |= operator executes analogously to how += does. It is equivalent to

quiz1 = quiz1 | 1UL << 27; // equivalent to quiz1 | = 1UL << 27;

Imagine that the teacher reexamined the quiz and discovered that student 27 actually had failed the test. The teacher must now turn off bit 27. This time we need an integer that has bit 27 turned off and all the other bits turned on. We’ll bitwise AND this value with quiz1 to turn off just that bit:

quiz1 &= ~(1UL << 27); // student number 27 failed

We obtain a value with all but bit 27 turned on by inverting our previous value. That value had 0 bits in all but bit 27, which was a 1. Applying the bitwise NOT to that value will turn off bit 27 and turn on all the others. When we bitwise AND this value with quiz1, all except bit 27 will remain unchanged.

Finally, we might want to know how the student at position 27 fared:

bool status = quiz1 & (1UL << 27); // how did student number 27 do?

Here we AND a value that has bit 27 turned on with quiz1. The result is nonzero (i.e., true) if bit 27 of quiz1 is also on; otherwise, it evaluates to zero.