C++ 初始化零数组
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Initializing an array of zeroes
提问by Yves Daoust
It is well known that missing initializers for an array of scalars are defaulted to zero.
众所周知，标量数组缺少的初始值设定项默认为零。
int A[5]; // Entries remain uninitialized
int B[5]= { 0 }; // All entries set to zero
But is this (below) guaranteed ?
但这（下面）有保证吗？
int C[5]= { }; // All entries set to zero
回答by Bathsheba
The empty braced initialisation performs aggregation-initialization of the array: this leads to zero-initialization of the intelements.
空括号初始化执行数组的聚合初始化：这导致int元素的零初始化。
Yes, this is guaranteed.
是的，这是有保证的。
回答by songyuanyao
Yes, according to the rule of aggregate initialization, it's guaranteed (that all elements of array Cwill be value-initialized, i.e. zero-initializedto 0in this case).
是，根据该规则集合初始化，它保证（即数组的所有元素C将值初始化，即零初始化到0在这种情况下）。
(emphasis mine)
（强调我的）
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 membersand bases (since C++17)are initializedby 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).
如果初始化子句的数量少于成员的数量
and bases (since C++17)或初始化列表完全为空，则其余成员由空列表and bases (since C++17)初始化by their default initializers, if provided in the class definition, and otherwise (since C++14)，按照通常的列表初始化规则（对非类类型执行值初始化和具有默认构造函数的非聚合类，以及聚合的聚合初始化）。
PS:
PS：
int A[5]; // Entries remain uninitialized
"remain uninitialized" might not be accurate. For int A[5];, all elements of Awill be default-initialized. If Ais static or thread-local object, the elements will be zero-initializedto 0, otherwise nothing is done, they'll be indeterminate values.
“保持未初始化”可能不准确。对于int A[5];， 的所有元素都A将被默认初始化。如果A是静态或线程本地对象，元素将被零初始化为0，否则什么都不做，它们将是不确定的值。
回答by rain_
In fact when you sayint A[5] = { 0 };you are saying: Initialize the first element to zero. All the other positions are initialized to zero because of the aggregate inizialization.
事实上，当你说int A[5] = { 0 };你是在说：将第一个元素初始化为零。由于聚合初始化，所有其他位置都初始化为零。
This line is the real responsible for having your array full of zeroes: int A[5] = { };
这一行真正负责让你的数组充满零： int A[5] = { };
That is why if you use int A[5] = { 1 };you will only have the first position inizialized to 1.
这就是为什么如果您使用，int A[5] = { 1 };您只会将第一个位置初始化为 1。
