问题描述
一个有趣的问题是此处其他问题的副作用，它涉及C和C ++处理方式（非静态存储持续时间）之间的可能差异：
An interesting question arose as a side effect of some other question here, about the possible differences between how C and C++ handle (the non-static-storage-duration):
int arr[7] = {0};
有人指出，在C ++中，其他元素不能保证为零，但我不确定我是否同意。
Someone was stating that, in C++, the other elements were not guaranteed to be zero but I'm not sure I agree.
现在C11状态为 6.7.9初始化/ 19 ：
这意味着 arr 的其他六个元素将被初始化为零（因为 static int x; 会将 x 初始化为零）。
This means that the other six elements of arr will be initilised to zero (since static int x; would initialise x to zero).
我不确定如果C ++也是如此。在C ++ 20标准中， 9.3.1聚合/ 3 指出：
I'm unsure if this is also the case for C++. In the C++20 standard, 9.3.1 Aggregates /3 states:
（3.1）-（与指定的初始化程序列表和类无关的东西-pax）。
(3.1) — (irrelevant stuff to do with designated initialiser lists and classes - pax).
（3.2）—如果初始化列表是一个初始化列表，则聚合的显式初始化元素是聚合的前 n 个元素，其中 n 是初始化列表中元素的数量。
(3.2) — If the initializer list is an initializer-list, the explicitly initialized elements of the aggregate are the first n elements of the aggregate, where n is the number of elements in the initializer list.
然后 / 4 说明显式初始化的工作方式，而 / 5 处理非显式情况：
Then /4 states how the explicit initialisations work, and /5 handles the non-explicit cases:
（5.1）—如果该元素有一个默认的成员初始化程序（10.3），该元素从该初始化程序初始化。
(5.1) — If the element has a default member initializer (10.3), the element is initialized from that initializer.
（5.2）—否则，如果该元素不是引用，则从空的初始化程序列表（9.3.4）。
(5.2) — Otherwise, if the element is not a reference, the element is copy-initialized from an empty initializer list (9.3.4).
（5.3）—否则，该程序格式不正确。
(5.3) — Otherwise, the program is ill-formed.
在我看来，我们的特殊情况包含在<$ c中$ c>（5.2），因此我们必须转到 9.3.4 来查看 int会发生什么初始化为空列表（ {} ）。这会经历很多情况，但我相信第一个匹配的是：
It appears to me that our particular case is covered by (5.2) so we have to go to 9.3.4 to see what happens to an int initialised with an empty list ({}). This goes through a lot of cases but I believe the first one that matches is:
然后，从 9.3初始化器/ 8 ：
（8.1）—如果T是a（可能是cv限定的）类类型（第10条），没有默认构造函数（10.3.4）或用户提供或删除的默认构造函数，则对象被默认初始化；
(8.1) — if T is a (possibly cv-qualified) class type (Clause 10) with either no default constructor (10.3.4) or a default constructor that is user-provided or deleted, then the object is default-initialized;
（8.2） —如果T是（可能是cv限定的）类类型，而没有用户提供或删除的默认构造函数，则将该对象初始化为零，并检查默认初始化的语义约束，以及T是否具有非平凡的默认值构造函数，对象是默认初始化的；
(8.2) — if T is a (possibly cv-qualified) class type without a user-provided or deleted default constructor, then the object is zero-initialized and the semantic constraints for default-initialization are checked, and if T has a non-trivial default constructor, the object is default-initialized;
（8.3）—如果T是数组类型，则每个元素都将被值初始化；
(8.3) — if T is an array type, then each element is value-initialized;
（8.4）—否则，对象是零初始化的。
(8.4) — otherwise, the object is zero-initialized.
因此，它是 8.4 ，它似乎是控制子句，意思是C ++也会 将数组的非显式元素初始化为零。
Hence, it's 8.4 which seems to be the controlling clause, meaning C++ will also initialise the non-explicit elements of the array to zero.
我的推理正确吗？ C ++在遇到 int arr [7] = {0}; 时，会将所有元素设置为零吗？
Is my reasoning correct? Will C++, on encountering int arr[7] = {0};, set all elements to zero?
推荐答案
是。 C ++通常保持与C的向后兼容性，从而允许您包含和使用C代码。考虑一下是否有一些遗留的C代码，它们会尝试像您描述的那样初始化数组：
Yes. C++ generally maintains backward compatibility with C, allowing you to include and use C code. Consider if you have some legacy C code, that attempts to initialize an array like you describe:
int arr [7] = {0};
如果C ++的工作方式有所不同，并且C程序（在C语言下有效）假定此数组初始化为零，则代码为如果包含在使用C ++编译器编译的C ++项目中，则很容易失败。
If C++ worked any differently, and the C program (validly, under C) assumed that this array is zero initialized, then the code is liable to fail if included in a C++ project compiled with a C++ compiler.
为了确认，我在x64 Windows上使用Cygwin g ++编译了此C ++程序：
Just to confirm, I compiled this C++ program using Cygwin g++ on x64 Windows:
int main() {
int arr[7] = {0};
}
然后反汇编函数 main 在GDB中：
and then disassembled function main in GDB:
push %rbp
mov %rsp,%rbp
sub $0x40,%rsp
callq 0x1004010d0 <__main>
movq $0x0,-0x20(%rbp)
movq $0x0,-0x18(%rbp)
movq $0x0,-0x10(%rbp)
movl $0x0,-0x8(%rbp)
mov $0x0,%eax
add $0x40,%rsp
pop %rbp
retq
如您所见，该程序将3个qword和1个dword的零值移入堆栈。这是28个字节，即我系统上7个整数的大小。
As you can see, the program moves 3 qwords and 1 dword worth of zeros into the stack. That's 28 bytes, which is the size of 7 ints on my system.
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