您可以使用任何需要常量积分表达式然后将被优化的表达式.

You can use any expression that needs a constant integral expression and that will then be optimized out.

#define NPOT(X)                                         \
 (1                                                     \
 ? complex_algorithm(X)                                 \
 : sizeof(struct { int needs_constant[1 ? 1 : (X)]; })  \
 )

最终，您应该将sizeof的结果强制转换为适当的整数类型，以便返回表达式具有您所期望的类型.

eventually you should cast the result of the sizeof to the appropriate integer type, so the return expression is of a type that you'd expect.

我在这里使用未加标签的struct

I am using an untagged struct here to

• 具有类型，因此实际上不会产生临时
• 具有独特的类型，使得表达式可以在代码中的任何地方重复而不会引起冲突
• 触发VLA的使用，从C99开始，在struct中不允许使用VLA:

• have a type so really no temporary is produced
• have a unique type such that the expression can be repeated anywhere in the code without causing conflicts
• trigger the use of a VLA, which is not allowed inside a struct as of C99:

结构或联合的成员可以具有除 可变修饰的类型.

A member of a structure or union may have any object type other than a variably modified type.

我正在将三元?:与1用作选择表达式，以确保始终对:的类型进行求值，而从未将其作为表达式求值.

I am using the ternary ?: with 1 as the selecting expression to ensure that the : is always evaluated for its type, but never evaluated as an expression.

似乎gcc接受struct内的VLA作为扩展名，甚至不警告它，即使我明确地说-std=c99也不行.对他们来说，这真是个坏主意.

It seems that gcc accepts VLA inside struct as an extension and doesn't even warn about it, even when I explicitly say -std=c99. This is really a bad idea of them.

对于这样一个奇怪的编译器:)，您可以改用sizeof((int[X]){ 0 }).与上述版本一样，这是被禁止的"，但是另外，甚至gcc也对此有所抱怨.

For such a weird compiler :) you could use sizeof((int[X]){ 0 }), instead. This is "as forbidden" as the above version, but additionally even gcc complains about it.