如何解决构造函数和类型转换运算符歧义的问题？
本帖最后由 shendaowu 于 2013-07-28 08:36:58 编辑

#include <iostream>

using namespace std;

class Test
{
public:
    Test ( int n );
    Test operator+( Test t );
    operator int();
    
    int m_n;
};

Test::Test( int n )
{
    cout << "Test(int) get" << n << endl;
    m_n = n;
}
Test Test::operator+( Test t )
{
    t.m_n += m_n;
    return t;
}
Test::operator int()
{
    cout << "operator int() from" << m_n << endl;
    return m_n;
}

ostream& operator<<( ostream& os, Test t )
{
    cout << t.m_n;
}

int main()
{
    Test t0( 1 );
    int s = 0;
    
    s = t0 + 1;
    cout << s << endl;
    
    return 0;
}

这个程序是有问题的。
错误：
Compiling the source code....
$g++ -std=c++11 main.cpp -o demo -lm -pthread -lgmpxx -lreadline 2>&1
main.cpp: In function 'int main()':
main.cpp:41:14: error: ambiguous overload for 'operator+' in 't0 + 1'
main.cpp:41:14: note: candidates are:
main.cpp:41:14: note: operator+(int, int)
main.cpp:20:6: note: Test Test::operator+(Test)

这个能解决么？是我什么地方搞错了么？

#include <iostream>

using namespace std;

class Test
{
public:
    explicit Test( int n );
    Test operator+( Test t );
    operator int();
    
    int m_n;
};

Test::Test( int n )
{
    cout << "Test(int) get " << n << endl;
    m_n = n; 
 
}
Test Test::operator+( Test t )
{
    t.m_n += m_n;
    return t;
}
Test::operator int()
{
    cout << "operator int() from " << m_n << endl;
    return m_n;
}

ostream& operator<<( ostream& os, Test t )
{
    cout << t.m_n;
}

int main()
{
    Test t0( 1 );
    int si = 0;
    Test st( 0 );
    
    si = t0 + 1;
    cout << si << endl;
    st = t0 + 1;
    cout << st << endl;
    
    return 0;
}

在树上看到explicit了，但是上面的还是会出错：
Compiling the source code....
$g++ -std=c++11 main.cpp -o demo -lm -pthread -lgmpxx -lreadline 2>&1
main.cpp: In function 'int main()':
main.cpp:44:15: error: no match for 'operator=' in 'st = (t0.Test::operator int() + 1)'
main.cpp:44:15: note: candidates are:
main.cpp:5:7: note: Test& Test::operator=(const Test&)
main.cpp:5:7: note: no known conversion for argument 1 from 'int' to 'const Test&'
main.cpp:5:7: note: Test& Test::operator=(Test&&)
main.cpp:5:7: note: no known conversion for argument 1 from 'int' to 'Test&&'


怎么设计才能同时保留这两个东西？
[解决办法]

#include <iostream>
 
using namespace std;
 
class Test
{
public:
    explicit Test( int n );
    Test operator+( Test t );
    operator int(); 
 
     
    int m_n;
};
 
Test::Test( int n )
{
    cout << "Test(int) get " << n << endl;
    m_n = n;
}
Test Test::operator+( Test t )
{
    t.m_n += m_n;
    return t;
}
Test::operator int()
{
    cout << "operator int() from " << m_n << endl;
    return m_n;
}
 
ostream& operator<<( ostream& os, Test t )
{
    cout << t.m_n;
}
 
int main()
{
    Test t0( 1 );
    int si = 0;
    Test st( 0 );
     
    si = (int)t0 + 1; 
    cout << si << endl;
    st = t0 + Test(1);//或者在写一个friend operator+（const T& t,const int& d)的重载
    cout << st << endl;
     
    return 0;
}