我正在尝试使用 GCC v4.3.x 设置静态断言(在主函数之外):
#define STATIC_ASSERT(cond) extern void static_assert(int arg[(cond) ? 1 : -1])
STATIC_ASSERT( (double)1 == (double)1 ); // failed
但是当我使用 float 时，断言总是失败。
是否可以正确运行此静态断言？
最佳答案
C++ 标准 2003，5.19“常量表达式”，第 1 段。
In several places, C++ requires expressions that evaluate to an integral or enumeration constant: as array bounds (8.3.4, 5.3.4), as case expressions (6.4.2), as bit-field lengths (9.6), as enumerator initializers (7.2), as static member initializers (9.4.2), and as integral or enumeration non-type template arguments (14.3).
constant-expression: conditional-expression
An integral constant-expression can involve only literals (2.13), enumerators, const variables or static data members of integral or enumeration types initialized with constant expressions (8.5), non-type tem- plate parameters of integral or enumeration types, and sizeof expressions. Floating literals (2.13.3) can appear only if they are cast to integral or enumeration types. Only type conversions to integral or enumeration types can be used. In particular, except in sizeof expressions, functions, class objects, pointers, or references shall not be used, and assignment, increment, decrement, function-call, or comma operators shall not be used.
关于c++ - 无效的静态断言行为，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1951741/