1.1.6.1 Implicit conversion in C++

• Compiler will first match function name and number of argument, then look for template deducted function, then use implicit conversion to try match. So implicit conversion happens after template.
• When will class implicit convert. 1) single ctor, 2) operator Type.
  class A { 
  A(int i); // bad 
  operator const char*(); //bad 
  }; 
   
  A a1, a2; 
  a2 = a1*2;  //implicit conversion 2 to temp A obj 
  a2 = 2  the same, then call operator =;
• You should always avoid it. 1) use explicit single ctor 2) use name convert function instead of "operator Type". so string has function c_str() instead operator char*() const. In example below, with explicit keyword before ctor, you have to use A(2) or (A)2 to explicitly build a A obj.
  class A { 
  explicit A(int i); // good 
  const char* getInternalPoint() ; 
  //good, use a name function. 
  };
• convert class to basic type, you need operator basicTypeName, no argument, no return value and member function.
• Implicit conversion can be called by compiler implicit, (means that you don’t know at all). It sometimes will lead to potential ambiguity problem.
  class A{ 
  A(class B&); 
  }; 
  class B{ 
  operator A() 
  }; 
   
  void g(const A&); 
  B b; 
  g(b) 
  // it can call A’s ctor in class A 
  // or it can call B’s opeartor A() in class B 
  //compiler will stop, it meet ambiguity.

  more detail can be seen effective c++ item 26.

• For class, assignment operator() can support two different type assignment. But I don’t think that it is a conversion.
  class A {}; 
   
  class B { 
  public: 
    // conversion from A (assignment): 
    B& operator= (const A& x) {return *this;} 
  }; 
   
    void fun(B x){}; 
    B b; 
    b = a;    // calls constructor