重写虚函数时的异常规范
考虑下面的代码：
class A
{
public:
virtual void f() throw ( int ) { }
};
class B: public A
{
public:
void f() throw ( int, double ) { }
};
编译时，它说派生类 B 与 A 相比具有更宽松的抛出说明符。这有什么重要性？如果我们尝试交换它们的异常规范，例如 A::f() 抛出 int 和 double，而 B::f() 只抛出 int，则不会出现错误。
Consider the following code:
class A
{
public:
virtual void f() throw ( int ) { }
};
class B: public A
{
public:
void f() throw ( int, double ) { }
};
When compiled, it says that derived class B has a looser throw specifier compared to A. What is the importance of this? If we try to exchange their exception specification, such that A::f() throws int and double while B::f() throws only int, the error does not appear.
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评论（2）
扩展第 2 点：
A的调用者期望只有int出来，但如果您使用B（其中，因为它是公开的）派生自A，也意味着它可以用作A），突然double也可以出现，这会破坏A< /code> 的契约（只有int被抛出）。To expand on point 2:
A's callers expect that onlyintcomes out, but if you use aB(which, because it's publicly derived fromA, also means it's usable as anA), suddenlydoublecan come out too, and that would breakA's contract (that onlyintgets thrown).你的 B 违反了里氏替换原则 - 例如：
void foo(A* a) throw() // ie I will not throw { try { a->f(); } catch(int) {} }根据 A 的接口，这是有效的；特别是，我们不希望抛出双倍。但考虑一下我们是否要
使用您描述的接口进行调用，并且
were to throw a double.
Your B violates the Liskov Substitution Principle - eg:
void foo(A* a) throw() // ie I will not throw { try { a->f(); } catch(int) {} }This is valid according to the interface for A; in particular, we do not expect a double to be thrown. But consider if we were to call
with the interface you describe, and
were to throw a double.