在 C++ 中组合初始化 class

Question

我不确定这是否叫compose initialization。基本上我想知道的是：

main 中的对象 foo 是如何初始化的
这种初始化有什么规则吗，因为我发现有时会失败。

给定以下代码：

#include <iostream>
#include <string>
#include <vector>

class Base {
public:
  std::string str_mem;
  int int_val = 0;
};

class Bar {
public:
  Bar() = default;
  Bar(int val) : int_val(val) {
  }

  auto GetVal() {
    return int_val;
  }

private:
  int int_val = 0;
};

class Foo : public Base {
public:
  int mem_1;
  int mem_2;
  int mem_3;

  std::string str_mem;
  Bar class_mem;

  auto print() {
    printf("[%d,%d,%d,%s,%d]\n", mem_1, mem_2, mem_3, str_mem.c_str(), class_mem.GetVal());
  }
};

auto main(int argc, char** argv) -> int {
  Foo foo{{"base1", 2}, 1, 2, 3}; // How `foo` is constructed, any constructor get called?
  foo.print();

  Foo foo1{{"base2", 2}, 1, 2, 3, "string", 3};
  foo1.print();
}

Answer 1

这是aggregate initialization, Foo is an aggregate type。

Each direct public base, (since C++17) array element, or non-static class member, in order of array subscript/appearance in the class definition, is copy-initialized from the corresponding clause of the initializer list.

给定 Foo foo1{{"base2", 2}, 1, 2, 3, "string", 3};，Base 的子对象从 {"base2", 2} 初始化（它也得到 aggregate-initialized），数据成员 mem_1 从1、mem_2从2初始化，mem_3从3初始化，str_mem从"string"初始化（通过std::string 的构造函数采用 const char*)，class_mem 从 3 初始化（由构造函数 Bar::Bar(int)）。

给定Foo foo{{"base1", 2}, 1, 2, 3};，除了str_mem和class_mem是从空列表初始化的，前面的成员的初始化方式与上面相同，结果是default-initialized（由它们的默认构造函数）。

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default member initializers, if provided in the class definition, and otherwise (since C++14) copy-initialized from empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates).

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