Using bitwise OR (|) with a mask to set multiple bits in an integer: is this a valid and common technique?

Introduction / Context:
Bit-setting with OR is one of the first mask techniques developers learn. It is used in flag fields, permissions, device registers, and protocol headers. The expression combines the existing bits with a mask where ones indicate which positions to turn on.

Given Data / Assumptions:

• x is an integer whose bits we want to modify.
• mask has 1s in the positions to set and 0s elsewhere.
• We perform x |= mask (or x = x | mask).

Concept / Approach:
OR has the truth table: 0|0=0, 0|1=1, 1|0=1, 1|1=1. Therefore, any position that is already 1 stays 1, and any position that is 0 becomes 1 when the mask has a 1 in that position. As a result, OR “sets” the specified bits and leaves others unchanged.

Step-by-Step Solution:

Verification / Alternative check:
Example: x = 0b0101, mask = 0b1010 → x | mask = 0b1111; both bits 1 and 3 are now set.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing OR for setting with AND-for-clearing; not using unsigned when building shift-based masks.

Final Answer:
Yes — x = x | mask (or x |= mask) sets all bits that are 1 in mask.