如何定义和使用带有“可选参数”的 boost::function ？
我正在使用一个需要某种回调方法的类，因此我使用 boost::function 来存储函数指针。
我需要回调有一个可选参数，但我发现 boost::function 不允许我定义可选参数类型，所以我尝试了以下代码并且它起作用了..
//the second argument is optional
typedef boost::function< int (int, char*)> myHandler;
class A
{
public:
//handler with 2 arguments
int foo(int x,char* a) {printf("%s\n",a); return 0;};
//handler with 1 argument
int boo(int x) {return 1;};
}
A* a = new A;
myHandler fooHandler= boost::bind(&A::foo,a,_1,_2);
myHandler booHandler= boost::bind(&A::boo,a,_1);
char* anyCharPtr = "just for demo";
//This works as expected calling a->foo(5,anyCharPtr)
fooHandler(5,anyCharPtr);
//Surprise, this also works as expected, calling a->boo(5) and ignores anyCharPtr
booHandler(5,anyCharPtr);
我很震惊它起作用了，问题它应该有效吗？它合法吗？
有更好的解决方案吗？
I am using a class that needs some kind of callback method, so i'm using boost::function to store the function pointers.
i need the callback to have one optional argument, but i found out that boost::function won't let me define optional arguments kind of type, so i tried the following code and it worked..
//the second argument is optional
typedef boost::function< int (int, char*)> myHandler;
class A
{
public:
//handler with 2 arguments
int foo(int x,char* a) {printf("%s\n",a); return 0;};
//handler with 1 argument
int boo(int x) {return 1;};
}
A* a = new A;
myHandler fooHandler= boost::bind(&A::foo,a,_1,_2);
myHandler booHandler= boost::bind(&A::boo,a,_1);
char* anyCharPtr = "just for demo";
//This works as expected calling a->foo(5,anyCharPtr)
fooHandler(5,anyCharPtr);
//Surprise, this also works as expected, calling a->boo(5) and ignores anyCharPtr
booHandler(5,anyCharPtr);
I was shocked that it worked, question is should it work, and is it legit?
is there a better solution?
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可以说是绑定中的类型安全漏洞 ->函数转换。
boost::bind 不返回 std::function 而是一个非常复杂类型的函数对象。在上面看到的情况下
，返回的对象具有类型
std::function 仅检查提供的函数对象是否“兼容”，在这种情况下，是否可以使用 int 调用 作为第一个参数，指向 char 的指针作为第二个参数。在检查 *boost::bind_t* 模板时，我们发现它确实有一个匹配的函数调用运算符：
在这个函数内，第二个参数最终被默默地丢弃。这是设计使然。来自文档：任何额外的参数都会被默默地忽略(...)
Arguably a type safety hole in the bind -> function conversion.
boost::bind doesn't return an std::function but a function object of a very complicated type. In the case of
as seen above, the returned object has type
std::function only checks that the supplied function object is "compatible", in this case, whether it is callable with an int as the first argument and a pointer to char as the second argument. Upon examining the *boost::bind_t* template, we see that it does indeed have a matching function call operator:
Inside this function the second argument ends up being silently discarded. This is by design. From the documentation: Any extra arguments are silently ignored (...)