C++：存在构造函数的结构初始化

Question

I have question on C++ behavior when initializing structures using a list. For example, the following code behaves the same in C and C++. The list initializes x :

struct X {
    int x;
};

int main(int argc, char *argv[])
{
    struct X xx = {0};    
    return 0;
}

Now, if I add a constructor, I find out through testing that the constructor is called instead of the simple initialization of the x member:

#include <iostream>
using namespace std;

struct X {
    int x;
    X(int);
};

X::X(int i)
{
    cout << "X(int i)" << endl;
    x = i;
}

int main(int argc, char *argv[])
{
    struct X xx = {0};

    return 0;
}

Output:

X(int i)

Is the behavior of C++ with an identical constructor (as above) to override the simple list initialization? Giving my testing that is what appears to happen.

Thanks!

Answer 1

The following syntax:

X xx = {0};

is just a form of copy-list initialization . This has the effect of invoking the constructor of X , as you observed.

The name of this initialization comes from the fact that this looks like a list is being copied, but just the regular constructor is invoked. Note that this will only consider implicit constructors.

Also the elaborated-type-specifier struct is not necessary in c++, in the declaration of xx .

Note that if you don't provide a constructor such as X(int) (as you did in the first example), then the declaration of xx will instead do aggregate initialization.