函数指针中的c++异常规范

Question

代码:

#include<iostream>
using namespace std;

void foo() throw(char) {throw 'a';}

int main() try {

   void (*pf)() throw(float);
   pf = foo; // This should NOT work 
   pf();

}
catch(const char& c){cout << "Catched ::> " << c << endl;}

为什么可以将 foo 传递给 pf，即使 foo 异常规范不同于函数指针 pf 有？这是我的编译器中的错误吗？

Answer 1

Exception specifications don’t participate in a function’s type. 更正:正如其他答案所指出的，这确实是一个编译器错误。众所周知，大多数编译器在实现异常规范时都存在缺陷。此外，它们在 C++11 中已弃用。所以，

遵循 Herb Sutter 的异常规范建议:

Moral #1: Never write an exception specification.

Moral #2: Except possibly an empty one, but if I were you I’d avoid even that.