我试图了解虚函数的工作原理,但我卡在了某个部分。
我写了这个小程序:
class First
{
public:
virtual void f(int a)
{
cout << "First!" << endl;
cout << a << endl;
}
};
class Second : public First
{
public:
void f(int a) {
cout << "Second!" << endl;
cout << a << endl;
}
};
void main() {
Second s;
First *p = &s;
p->f(5);
First n;
p = &n;
p->f(3);
_getch();
}
此代码导致:
Second! 5 First! 3
However, if I change int
in the Second::f()
function to a different type, like this:
class First
{
public:
virtual void f(int a) {
cout << "First!" << endl;
cout << a << endl;
}
};
class Second : public First
{
public:
void f(double a) { //double instead int here!
cout << "Second!" << endl;
cout << a << endl;
}
};
void main() {
Second s;
First *p = &s;
p->f(5);
First n;
p = &n;
p->f(3);
_getch();
}
我的程序从不调用 Second::f()
,结果是这样的:
First! 5 First! 3
有人可以向我解释为什么会这样吗?
最佳答案
当使用虚函数调度时,所谓的“最终覆盖器”就是被调用的。对于甚至覆盖继承的虚函数的函数,它必须满足一些条件:
If a virtual member function
vf
is declared in a classBase
and in a classDerived
, derived directly or indirectly fromBase
, a member functionvf
with the same name, parameter-type-list (8.3.5), cv-qualification, and refqualifier (or absence of same) asBase::vf
is declared, thenDerived::vf
is also virtual (whether or not it is so declared) and it overridesBase::vf
.
-- ISO/IEC 14882:2001(E) §10.3(粗体强调我的)
很简单,在您的第二个示例中,Second::f(double)
的参数列表与 First::f(int)
的参数列表不同,因此 Second::f(double)
是不是(自动)虚拟的并且不是覆盖First::f(int)
.
C++11 关键字 override
声明了您的意图,即方法覆盖继承的虚拟方法,以便编译器可以在不覆盖时告诉您。例如,如果您改为这样做:
void f(double a) override {
编译器会给你这个诊断来告诉你它实际上并没有覆盖任何东西,它甚至会告诉你为什么它没有(“第一个参数类型不匹配('int' vs 'double ')"):
main.cpp:15:18: error: non-virtual member function marked 'override' hides virtual member function
void f(double a) override { //double instead int here!
^
main.cpp:7:14: note: hidden overloaded virtual function 'First::f' declared here: type mismatch at 1st parameter ('int' vs 'double')
virtual void f(int a) {
^
关于c++ - 具有不同参数类型的虚函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43644194/