使用结构名称作为函数

Question

I'm trying to understand what the following line does:

BStats stats = BStats();

The struct is defined as follows:

struct BStats
{
    unsigned a;
    unsigned b;

    BStats& operator+=(const BStats& rhs)
    {
        this->a += rhs.a;
        this->b += rhs.b;
        return *this;
    }
};

But I have no idea about what this line does. Is it calling the default constructor?

Answer 1

The expression BStats() is described in the standard in 5.2.3/2:

The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, which is value-initialized.

That is, the expression creates an rvalue of Bstats type that is value-initialized . In your particular case, value-initialization means that the two members of the BStats struct will be set to zero.

Note that this is different than the behavior of calling the default-constructor that is mentioned in other answers, as the default constructor will not guarantee that the members are set to 0.

Answer 2

Just like any class, a struct has a default constructor automatically created by the compiler. In your case, BStats() simply calls the default constructor, although the explicit call is useless.

Answer 3

在C ++中，类和结构几乎相同（不同之处在于C ++结构是具有public作为默认属性的类，而类的属性是私有的），就像调用构造函数一样。