c++ - g++ 6.2 的不同异常说明符

Question

有人能给我解释一下为什么这段代码不能用 g++ 版本 6.2.0 编译，但是可以用 clang++ 版本 3.9.0-svn274438-1 和 icpc 版本 16.0.2 编译吗

$ cat wtf.cpp
#include <cstdio>
#include <new>
void *operator new(std::size_t) throw(std::bad_alloc);
void *operator new(std::size_t) throw (std::bad_alloc) { void *p; return p; }

$ g++-6 wtf.cpp -c 
wtf.cpp: In function ‘void* operator new(std::size_t)’:
wtf.cpp:4:7: error: declaration of ‘void* operator new(std::size_t) throw (std::bad_alloc)’ has a different exception specifier
 void *operator new(std::size_t) throw (std::bad_alloc) { void * p; return p; }
       ^~~~~~~~
wtf.cpp:3:7: note: from previous declaration ‘void* operator new(std::size_t)’
 void *operator new(std::size_t) throw(std::bad_alloc);

Answer 1

您使用的是 C++11 或更高版本吗？

C++98 中原始的 operator new() 声明

throwing:   
void* operator new (std::size_t size) throw (std::bad_alloc);

nothrow:
void* operator new (std::size_t size, const std::nothrow_t& nothrow_value) throw();

placement:
void* operator new (std::size_t size, void* ptr) throw();

已在 C++11 中更改为使用 noexcept 关键字:

throwing:   
void* operator new (std::size_t size);

nothrow:    
void* operator new (std::size_t size, const std::nothrow_t& nothrow_value) noexcept;

placement:  
void* operator new (std::size_t size, void* ptr) noexcept;

Reference link.