C++ Programming :: question about auto


	question about auto_ptr initialization int p = new int; auto_ptr<int> ap1 = p; //will fail on 3 1 auto_ptr<int> ap1 p ; //ok 2 ap1 = 12; // 3 the first situation has called : explicit auto_ptr(_Ty _Ptr = 0) _THROW0() the second : auto_ptr(auto_ptr_ref<_Ty> _Right) _THROW0() why the two expression have called the different ctor? is the 1 expression has compiled as follow? auto_ptr<int> ap1; ap1 = p; On 5 26 , 10 53 , timlyee <yiyem@gmail.com> wrote: > int p = new int; > auto_ptr<int> ap1 = p; //will fail on 3 1 > auto_ptr<int> ap1 p ; //ok 2 > ap1 = 12; // 3 > the first situation has called : explicit auto_ptr(_Ty _Ptr = 0) > _THROW0() > the second : auto_ptr(auto_ptr_ref<_Ty> _Right) _THROW0() > why the two expression have called the different ctor? Are you sure you can compile above code correctly? When I try to compile the above code, the compiler(MinGW gcc 4.1.2) said "error: conversion from 'int' to non-scalar type 'std::auto_ptr<int>' requested" at "auto_ptr<int> ap1 = p;". > is the 1 expression has compiled as follow? > auto_ptr<int> ap1; > ap1 = p; I think so. And I remember that auto_ptr doesn't overload the '=' operator for a pointer. On May 26, 4:53 am, timlyee <yiyem@gmail.com> wrote: > int p = new int; > auto_ptr<int> ap1 = p; //will fail on 3 1 > auto_ptr<int> ap1(p); //ok 2 > ap1 = 12; // 3 > the first situation has called : explicit auto_ptr(_Ty _Ptr = 0) > _THROW0() According to the standard, the line labeled 1 first converts the argument to the target type, then copies. Since the one argument constructor is marked explicit, it can't be used for conversion; in fact, there is no implict conversion, and line one should fail to compile. (G++ says: "error: conversion from 'int' to non-scalar type 'std::auto_ptr<int>' requested") Many of the earlier versions of auto_ptr allowed it, so if you have an older compler, it might compile; some newer ones might allow it as well, to avoid breaking older code. > the second : auto_ptr(auto_ptr_ref<_Ty> _Right) _THROW0() Nothing in your code should call this. The line labeled 2 calls st::auto_ptr<int>( int ), exactly as it looks like. In this case, no conversion is involved, so the "explicit" doesn't inhibit anything. > why the two expression have called the different ctor? Because they have different semantics. The statement labeled 1 is "copy initialization", and the statement labeled 2 is "direct initialization". Although the exact details are a bit more complex, the basic principle is that copy initialization converts the initialization to the target type, then uses the copy constructor to copy it. (The additional temporary, and thus the copy constructor, may be elided, but the legality of the code may not depend on this elision.) Direct initialization does exactly what it's name says, it calls the constructor for the object with the given arguments. > is the 1 expression has compiled as follow? > auto_ptr<int> ap1; > ap1 = p; Never. Despite the =, copy initialization is not assignment, and does not involve assignment. (Of course, this won't compile with auto_ptr either, since there is no assignment operator which takes a pointer. But in earlier versions, there was, so your milage may vary.) -- James Kanze (Gabi Software) email: james.ka@gmail.com Conseils en informatique oriente objet/ Beratung in objektorientierter Datenverarbeitung 9 place Smard, 78210 St.-Cyr-l'cole, France, +33 (0)1 30 23 00 34

C++ Programming

question about auto_ptr initialization