Does C 's Implicit Default Constructor Initialize Built-in Type Members?

Initialization of Built-in Types by the Default Constructor

In C , the default constructor is implicitly defined by the compiler for classes that do not have an explicitly declared constructor. Does this implicit default constructor automatically initialize members of built-in types?

Answer

No, the implicitly defined default constructor does not initialize members of built-in types. However, it's important to note that class instances can be initialized in other ways.

Value-Initialization vs. Default Constructor

Commonly, the syntax C() is assumed to invoke the default constructor. However, in certain cases, it performs value-initialization instead. This occurs if a user-declared default constructor doesn't exist. Value-initialization directly initializes each class member, resulting in zero-initialization for built-in types.

For example:

class C {
public:
  int x;
};

If no user-declared constructor is defined, C() will use value-initialization:

C c;
// c.x contains garbage

Explicit Value-Initialization

Explicit value-initialization using (), as seen in the following code, will zero-initialize x:

C c = C();
// c.x == 0

C *pc = new C();
// pc->x == 0

Aggregate Initialization

Aggregate initialization can also initialize class members without using a constructor:

C c = {}; // C++98
C d{}; // C++11
// c.x == 0, d.x == 0

Therefore, while the default constructor does not initialize built-in member types, alternative initialization methods exist in C .

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