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。

当调用构造函数时，它实际上是从堆栈或堆中分配内存。因此，在另一个构造函数中调用一个构造函数会创建一个本地副本。所以我们正在修改另一个对象，而不是我们所关注的对象。

c++ 11:是的!

c++ 11及以后的版本也有同样的特性(称为委托构造函数)。

语法与c#略有不同:

class Foo {
public: 
  Foo(char x, int y) {}
  Foo(int y) : Foo('a', y) {}
};

c++ 03:不

不幸的是，在c++ 03中没有办法做到这一点，但是有两种方法来模拟这个:

你可以通过默认参数组合两个(或更多)构造函数: 类Foo { 公众: Foo(char x, int y=0);//结合两个构造函数(char)和(char, int) / /…… }; 使用init方法共享公共代码: 类Foo { 公众: Foo (char x); Foo(char x, int y); / /…… 私人: Void init(char x, int y); }; Foo:: Foo (char x) ｛ Init (x, int(x) + 7); / /…… ｝ Foo::Foo(char x, int y) ｛ init (x, y); / /…… ｝ void Foo::init(char x, int y) ｛ / /…… ｝

请参阅c++ FAQ条目以获得参考。

在c++ 11中，一个构造函数可以调用另一个构造函数重载:

class Foo  {
     int d;         
public:
    Foo  (int i) : d(i) {}
    Foo  () : Foo(42) {} //New to C++11
};

此外，成员也可以这样初始化。

class Foo  {
     int d = 5;         
public:
    Foo  (int i) : d(i) {}
};

这样就不需要创建初始化帮助器方法了。并且仍然建议不要在构造函数或析构函数中调用任何虚函数，以避免使用任何可能未初始化的成员。

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:)