C++可变参数函数指针
以下是否会产生明确的行为？也就是说，如果将非可变参数函数 f 转换为可变参数函数 g 并使用 f 期望的参数调用 g，则其行为是否与使用这些参数调用 f 的行为相匹配？
class Base {};
class Derived1 : public Base {
public:
int getInt1() {return 1;}
};
class Derived2 : public Base {
public:
int getInt2() {return 2;}
};
typedef int (*vfunc)(...);
int foo (vfunc f) {
Derived1 d1;
Derived2 d2;
return f(&d1, &d2);
}
int bar (Derived1 * p1, Derived2 * p2) {
return p1->getInt1() + p2->getInt2();
}
int main (int argc, char ** argv) {
return foo((vfunc)bar); // Is this program guaranteed to return 3?
}
更新
有什么方法可以让程序得到明确的定义，即使使用专有关键字也是如此？例如做一些类似这里提到的 __cdecl 的事情：
http://msdn.microsoft.com/en-us/library/984x0h58%28v=vs.80%29.aspx
我的最终目标是拥有一个尝试匹配 X 指针列表的 matcher 函数。匹配器函数接受一个谓词（不一定是一个函数......可能是一个列表）并接受一个将匹配结果传递给的函数。传递给它的回调函数采用与匹配的谓词相同的参数类型和数量。
Does the following result in well-defined behavior? That is, if you cast a non-vararg function f as a vararg function g and call g with the arguments that f expects, does the behavior match that of calling f with those arguments?
class Base {};
class Derived1 : public Base {
public:
int getInt1() {return 1;}
};
class Derived2 : public Base {
public:
int getInt2() {return 2;}
};
typedef int (*vfunc)(...);
int foo (vfunc f) {
Derived1 d1;
Derived2 d2;
return f(&d1, &d2);
}
int bar (Derived1 * p1, Derived2 * p2) {
return p1->getInt1() + p2->getInt2();
}
int main (int argc, char ** argv) {
return foo((vfunc)bar); // Is this program guaranteed to return 3?
}
UPDATE
Is there some way I can get the program to be well-defined, even if using proprietary keywords? Such as doing some stuff like __cdecl mentioned here:
http://msdn.microsoft.com/en-us/library/984x0h58%28v=vs.80%29.aspx
My end goal is to have a matcher function that tries matching on a list of X pointers. The matcher function takes in a predicate (not necessarily a function... might be a list) and takes in a function that it will pass the matched results to. The callback function passed to it takes the same argument types and arity as the predicate matched.
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否，根据 C++11 5.2.11/6 (
reinterpret_cast)，该行为未定义：bar的类型为int(Derived1*, Derived2*)。f（进行调用的表达式）指向的函数类型是int(...)。两者不同，因此行为未定义。No, the behavior is undefined, per C++11 5.2.11/6 (
reinterpret_cast):The type of
barisint(Derived1*, Derived2*). The type of the function pointed to byf(the expression through which the call is made) isint(...). The two are not the same and therefore the behavior is undefined.我很确定答案是“不”。
例如，在 Visual C++ 中，可变参数函数将具有与普通函数不同的调用约定（使用
/Gz时）。调用约定决定了生成什么预调用和后调用汇编代码，并且您不能安全地混合这两者。
I'm pretty sure the answer is "No."
For instance, in Visual C++, a variadic function will have a different calling convention than a normal function (when using
/Gz).The calling convention determines what pre-call and post-call assembly code is generated, and you cannot safely mix the two.