根据 this
void operator delete (void*); (1)
void operator delete[](void*); (2)
void operator delete (void*, const std::nothrow_t&); (3)
void operator delete[](void*, const std::nothrow_t&); (4)
void operator delete(void*, std::size_t) (5)
void operator delete[](void*, std::size_t) (6)
void operator delete(void*, std::size_t, const std::nothrow_t&) (7)
void operator delete[](void*, std::size_t, const std::nothrow_t&) (8)
(5-8) Called instead of (1-4) if a user-defined replacement is provided. The standard library implementations are identical to (1-4).
我相信引用的陈述是正确的；但是，我无法根据 C++14 标准草案 n3797 确认它。
我检查了 3.7.4 和 18.6，没有发现任何明确要求删除表达式必须调用 void::operator delete(void*, std::size_t) 而不是 void : :operator delete(void*) 如果前者存在。
你能给我指一下标准草案的右页吗？
最佳答案
释放函数的选择解释如下:
5.3.5 删除[expr.delete]
10 If the type is complete and if deallocation function lookup finds both a usual deallocation function with only a pointer parameter and a usual deallocation function with both a pointer parameter and a size parameter, then the selected deallocation function shall be the one with two parameters. Otherwise, the selected deallocation function shall be the function with one parameter.
关于c++ - C++14 是否要求删除表达式必须调用 `void operator::delete(void*, std::size_t)` 而不是 `void::operator delete(void*)` ？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24603894/