C ++初始化器列表的元素少于struct [英] C++ initializer lists with fewer elements than the struct

问题描述

当我使用初始化列表创建一个结构，但初始化列表包含比我的结构更少的元素，我看到剩余的元素用零初始化。

When I use an initializer list to create a struct, but the initializer list contains fewer elements than my struct, I see the remaining elements are initialized with zeroes.

这是一个未定义的行为，我看到零，因为我的编译器（VS2015）决定为我的内存零。

Is this an undefined behaviour and I'm seeing zeroes because my compiler (VS2015) decided to zero the memory for me?

或者有人可能指向我的文档解释这个行为在C ++中？

Or could someone point me to the documentation that explains this behaviour in C++?

这是我的代码：

struct Thing {
    int value;
    int* ptr;
};


void main() {
    Thing thing { 5 };
    std::cout << thing.value << " " << thing.ptr << std::endl;
}

这是它打印的：

5 00000000

推荐答案

这是一个定义的行为。根据聚合初始化的规则，剩余成员 ptr 将值初始化，即零初始化为NULL指针。

This is defined behaviour. According to the rule of aggregate initialization, the remaining member ptr will be value-initialized, i.e. zero-initialized to NULL pointer.

（强调我的）

如果initializer子句的数目小于 （因为C ++ 17）或初始化器列表是完全空的，剩余的成员和基数（自C ++ 17以来） c $ c>通过它们的默认初始化器，如果在类定义中提供，否则（由于C ++ 14）由空列表，根据通常的列表初始化规则>具有默认构造函数的非类类型和非聚合类的值初始化，以及聚合的聚合初始化）。如果引用类型的成员是这些剩余成员之一，则该程序是错误的。

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 initializers, if provided in the class definition, and otherwise (since C++14) by 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). If a member of a reference type is one of these remaining members, the program is ill-formed.

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