编译器在程序代码中提供私有构造函数作为候选

Question

#include <iostream>

using namespace std;

class Test{
private:
    Test(int a, int b=0)
    {
        cout << "private constructor\n";
    }
public:
    Test(int a)
    {
        cout << "public constructor\n";
    }
};

int main()
{
    Test t(1);
}

test.cpp: In function ‘int main()’:
test.cpp:20:10: error: call of overloaded ‘Test(int)’ is ambiguous
  Test t(1);
          ^
test.cpp:12:2: note: candidate: Test::Test(int)
  Test(int a)
  ^
test.cpp:7:2: note: candidate: Test::Test(int, int)
  Test(int a, int b=0)
  ^
test.cpp:5:7: note: candidate: Test::Test(const Test&)
 class Test{
       ^

test.cpp:20:7: error: call to constructor of 'Test' is ambiguous
        Test t(1);
             ^ ~
test.cpp:7:2: note: candidate constructor
        Test(int a, int b=0)
        ^
test.cpp:12:2: note: candidate constructor
        Test(int a)
        ^
test.cpp:5:7: note: candidate is the implicit copy constructor
class Test{
      ^
1 error generated.

Answer 1

根据标准，由于明确的原因，它是模棱两可的：

[class.access/5]

应该注意的是，受控制的是对成员和基类的访问，而不是它们的可见性。成员的名称仍然可见，当这些成员和基类不可访问时，仍然考虑对基类的隐式转换。对给定构造的解释是不考虑访问控制而建立的。如果建立的解释使用不可访问的成员名称或基类，则构造格式不正确。

int const default_b = 0;

// ...

return Test{a, default_b};