作为一个c#开发人员,我习惯了遍历构造函数:
class Test {
public Test() {
DoSomething();
}
public Test(int count) : this() {
DoSomethingWithCount(count);
}
public Test(int count, string name) : this(count) {
DoSomethingWithName(name);
}
}
在c++中有办法做到这一点吗?
我尝试调用类名和使用'this'关键字,但都失败了。
简单地说,你不能在c++ 11之前。
c++ 11引入了委托构造函数:
Delegating constructor
If the name of the class itself appears as class-or-identifier in the
member initializer list, then the list must consist of that one member
initializer only; such constructor is known as the delegating
constructor, and the constructor selected by the only member of the
initializer list is the target constructor
In this case, the target constructor is selected by overload
resolution and executed first, then the control returns to the
delegating constructor and its body is executed.
Delegating constructors cannot be recursive.
class Foo {
public:
Foo(char x, int y) {}
Foo(int y) : Foo('a', y) {} // Foo(int) delegates to Foo(char,int)
};
请注意,委托构造函数是一个全有或全无的提议;如果一个构造函数委托给另一个构造函数,则调用构造函数的初始化列表中不允许有任何其他成员。如果只初始化一次const/reference成员,这是有意义的。
测试比决定更容易:)
试试这个:
#include <iostream>
class A {
public:
A( int a) : m_a(a) {
std::cout << "A::Ctor" << std::endl;
}
~A() {
std::cout << "A::dtor" << std::endl;
}
public:
int m_a;
};
class B : public A {
public:
B( int a, int b) : m_b(b), A(a) {}
public:
int m_b;
};
int main() {
B b(9, 6);
std::cout << "Test constructor delegation a = " << b.m_a << "; b = " << b.m_b << std::endl;
return 0;
}
然后用98 std编译:
g++ main.cpp -std=c++98 -o test_1
你会看到:
A::Ctor
Test constructor delegation a = 9; b = 6
A::dtor
所以:)
在Visual c++中,你也可以在构造函数中使用这个符号:this->Classname::Classname(另一个构造函数的参数)。请看下面的例子:
class Vertex
{
private:
int x, y;
public:
Vertex(int xCoo, int yCoo): x(xCoo), y(yCoo) {}
Vertex()
{
this->Vertex::Vertex(-1, -1);
}
};
我不知道它在其他地方是否可以工作,我只在Visual c++ 2003和2008中测试过。我想,您也可以这样调用几个构造函数,就像在Java和c#中一样。
附言:坦率地说,我很惊讶之前没有提到这一点。