为何Linux内核中部分枚举的首个成员显式初始化为0？

Great question! Even though every C standard since ANSI C has specified that the first enumerator in an enum defaults to 0, Linux kernel developers often choose to explicitly set it to 0 anyway. Here are the key reasons:

• Self-documenting code for clarity
  Not every developer working on the kernel might have the C standard's default enumerator rules memorized, especially newer contributors. Explicitly writing = 0 removes any ambiguity—anyone reading the code can immediately see the value of the first member without having to recall language specs. This makes the code more accessible and reduces cognitive load.

• Defensive programming against future changes
  Suppose someone later modifies the enum by adding a new member at the start. If the original first member didn't have an explicit = 0, the new member would take that default value, shifting all existing members' values up by one. By explicitly setting the original first member to 0, you lock in its value—adding a layer of protection for the intended behavior of that key member, even if the enum's structure gets adjusted later.

• Alignment with kernel coding style
  The Linux kernel prioritizes consistency and explicitness in code. If other members in the enum have explicit values (or if the codebase generally favors explicit initializations), setting the first member to 0 keeps the style uniform. This makes the entire enum definition look intentional and polished, rather than relying on implicit language behavior.

• Emphasizing semantic importance
  Often, the first enumerator represents a default, fallback, or "base" case—like I2C_ADAPTER_SMBUS in your example. Explicitly initializing it to 0 draws attention to its special role, signaling to readers that this value isn't just a side effect of the C standard, but a deliberate choice tied to the enum's purpose.

内容的提问来源于stack exchange，提问作者Lazureus