C ++默认构造函数

Question

Given the following code:

class temp
{
public:
    string str;
    int num;
};

int main()
{
    temp temp1;
    temp temp2 = temp();

    cout << temp1.str << endl; //Print ""
    cout << temp2.str << endl; //Print ""

    cout << temp1.num << endl; //Print a rand num
    cout << temp2.num << endl; //Print 0
}

What is the different between these two?—

temp temp1;

and

temp temp2 = temp();

Answer 1

temp temp1;

This calls temp 's default constructor on the instance called temp1 .

temp temp2 = temp();

This calls temp 's default constructor on a temporary object, then calls the compiler-generated copy-constructor on temp2 with the temporary object as the argument (this of course assumes that the compiler doesn't elide copies; it depends on your compiler's optimization settings).

As for why you get different initialized values, section 8.5 of the standard is relevant:

---

8.5 Initializers [dcl.init]

Paragraph 5:

To zero-initialize an object of type T means:

• if T is a scalar type (3.9), the object is set to the value of 0 (zero) converted to T ;
• if T is a non-union class type, each nonstatic data member and each base-class subobject is zero-initialized;
• if T is a union type, the object's first named data member is zero-initialized;
• if T is an array type, each element is zero-initialized;
• if T is a reference type, no initialization is performed.

To default-initialize an object of type T means:

• if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
• if T is an array type, each element is default-initialized;
• otherwise, the object is zero-initialized.

To value-initialize an object of type T means:

• if T is a class type (clause 9) with a user-declared constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
• if T is a non-union class type without a user-declared constructor, then every non-static data member and base-class component of T is value-initialized;
• if T is an array type, then each element is value-initialized;
• otherwise, the object is zero-initialized.

Paragraph 7:

An object whose initializer is an empty set of parentheses, ie, (), shall be value-initialized.

Paragraph 9:

If no initializer is specified for an object, and the object is of (possibly cv-qualified) non-POD class type (or array thereof), the object shall be default-initialized; if the object is of const-qualified type, the underlying class type shall have a user-declared default constructor. Otherwise, if no initializer is specified for a nonstatic object, the object and its subobjects, if any, have an indeterminate initial value; if the object or any of its subobjects are of const-qualified type, the program is ill-formed.

12 Special Member Functions [special]

Paragraph 7:

An implicitly-declared default constructor for a class is implicitly defined when it is used to create an object of its class type (1.8). The implicitly-defined default constructor performs the set of initializations of the class that would be performed by a user-written default constructor for that class with an empty mem-initializer-list (12.6.2) and an empty function body.

12.6.2 Initializing bases and members [class.base.init]

Paragraph 4:

If a given nonstatic data member or base class is not named by a mem-initializer-id (including the case where there is no mem-initializer-list because the constructor has no ctor-initializer), then

• If the entity is a nonstatic data member of (possibly cv-qualified) class type (or array thereof) or a base class, and the entity class is a non-POD class, the entity is default-initialized (8.5). If the entity is a nonstatic data member of a const-qualified type, the entity class shall have a user-declared default constructor.
• Otherwise, the entity is not initialized. If the entity is of const-qualified type or reference type, or of a (possibly cv-qualified) POD class type (or array thereof) containing (directly or indirectly) a member of a const-qualified type, the program is ill-formed.

---

So now that the rules have been laid out, let's see how they apply:

 temp temp1; 

temp is a non-POD type (because it has a std::string member), and since no initializer is specified for temp1 , it will be default-initialized (8.5/9). This calls the default constructor (8.5/5). temp has an implicit default constructor (12/7) which default-initializes the std::string member and the int member isn't initialized at all (12.6.2/4).

 temp temp2 = temp(); 

On the other hand, the temporary temp object is value-initialized (8.5/7), which value-initializes all data members (8.5/5), which calls the default constructor in the std::string member and zero-initializes the int member (8.5/5).

Of course, if you much rather not have to refer to the standard in 5+ different places, just ensure that you explicitly initialize everything (eg int i = 0; or using initializer lists).

Answer 2

The behavior of your code depends critically on the compiler you are using. More precisely, it depends on which version of language specification your compiler implements.

For C++98 compilers, both declarations have identical effect on the final values of the objects being declared: the str member should become empty, while the num members should contain unpredictable value. In both cases the actual initialization is default-initialization performed by a compiler-provided default constructor of class temp . That default constructor initializes str , but leaves num uninitialized.

For C++03 compilers the behavior is different. There's no difference for temp1 object (its num is still unpredictable). But temp2 initialization is handled differently. In C++03 the () initializer triggers the new kind of initialization - so called value-initialization . Value-initialization ignores the compiler-provided default constructor of the top level object, and instead works directly on its subobjects (data members in this case). So the temp2 object is effectively initialized by value-initialization, which also sets the num member to zero (in addition to initializing str with an empty string). For this reason, temp2.num ends up being zero in C++03 compilers.

If in your experiments you observed consistent zero in temp2.num , it means that your compiler follows the C++03 specification in this respect.

Answer 3

temp temp1;

Will create a default initialized temp object. Since you provided no default constructor for temp , every member of temp will be default initialized too. Since std::string provides a default ctor, it gets initialized correctly and has a well-defined value. The integer, however, gets default initialized, which is implementation defined and normally a random value.

temp temp2 = temp();

This will first create a value initialized temp object. This is important, because the object itself is value initialized, so are its members. It doesn't matter for the string, as default and value initialization are the same, but it matters for the integer. A value initialized integer has the value 0 .
After that, you just copy over those members into temp2 .

Also, this relevant question might be of interest to you.
Edit : See my comment on @In silico's answer for explanation on why this isn't the case for MSVC. :/