枚举值定义中的逗号？

Question

我有以下标志枚举定义(使用 Cocoa 宏):

typedef NS_OPTIONS(uint64_t, ItemDataType) {

...

    A = (1 << 16),
    B = (1 << 17),
    C = (1 << 18),
...

    G = (1 << 31),

    TEST = (A | B | C, G)
};

TEST 应该是 (A | B | C | G) 但我们打错了，输入了 (A | B | C, G) 。这个符号是什么意思？ TEST 的实际值是多少？这不是编译器错误吗？

Answer 1

, 是 comma operator ，它是有效代码，它将计算左操作数，然后丢弃该值并计算右操作数，这将是结果。

来自 C99 草案标准部分 6.5.17 Comma operator paragraph 2:

The left operand of a comma operator is evaluated as a void expression; there is a sequence point after its evaluation. Then the right operand is evaluated; the result has its type and value.⁹⁷⁾

还值得注意的是，逗号运算符具有 lowest operator precedence .

如果您使用 -Weverything 标志，

clang 会提供以下警告:

warning: expression is not an integer constant expression; folding it to a constant is a GNU extension [-Wgnu-folding-constant]
TEST = (A | B | C , G)
       ^~~~~~~~~~~~~~~

正确的值必须是常量表达式，但事实并非如此，因为常量表达式不能包含逗号运算符。

gcc 在这种情况下会给我一个错误，但如果我使用 -Wall 标志也会发出警告:

warning: left-hand operand of comma expression has no effect [-Wunused-value]
 TEST = (A | B | C , G)
                   ^
error: enumerator value for 'TEST' is not an integer constant