我已经模板化了 gray_code
类,该类旨在存储一些无符号整数,其基础位以格雷码顺序存储。这里是:
template<typename UnsignedInt>
struct gray_code
{
static_assert(std::is_unsigned<UnsignedInt>::value,
"gray code only supports built-in unsigned integers");
// Variable containing the gray code
UnsignedInt value;
// Default constructor
constexpr gray_code()
= default;
// Construction from UnsignedInt
constexpr explicit gray_code(UnsignedInt value):
value( (value >> 1) ^ value )
{}
// Other methods...
};
在一些通用算法中,我写了这样的东西:
template<typename UnsignedInt>
void foo( /* ... */ )
{
gray_code<UnsignedInt> bar{};
// Other stuff...
}
在这段代码中,我希望 bar
初始化为零,因此 bar.value
初始化为零。然而,在与意想不到的错误作斗争之后,似乎 bar.value
被初始化为垃圾(准确地说是 4606858)而不是 0u
。这让我很吃惊,所以我去 cppreference.com 看看上面那行到底应该做什么......
据我所知,T object{};
形式对应于 value initialization .我觉得这句话很有趣:
In all cases, if the empty pair of braces {} is used and T is an aggregate type, aggregate-initialization is performed instead of value-initialization.
但是,gray_code
有一个用户提供的构造函数。因此它不是一个集合,因此 aggregate initialization不执行。 gray_code
没有构造函数采用 std::initializer_list
所以 list initialization也不执行。 gray_code
的值初始化应该遵循通常的 C++14 值初始化规则:
1) If T is a class type with no default constructor or with a user-provided default constructor or with a deleted default constructor, the object is default-initialized.
2) If T is a class type without a user-provided or deleted default constructor (that is, it may be a class with a defaulted default constructor or with an implicitly-defined one) then the object is zero-initialized and then it is default-initialized if it has a non-trivial default constructor.
3) If T is an array type, each element of the array is value-initialized.
4) Otherwise, the object is zero-initialized.
如果我没看错,gray_code
有一个显式默认的(不是用户提供的)默认构造函数,因此 1) 不适用。它有一个默认的默认构造函数,所以 2) 适用:gray_code
是 zero-initialized .默认的默认构造函数似乎满足了普通默认构造函数的所有要求,因此不应发生默认初始化。那么我们来看看gray_code
是如何被零初始化的:
If T is a scalar type, the object's initial value is the integral constant zero implicitly converted to T.
If T is an non-union class type, all base classes and non-static data members are zero-initialized, and all padding is initialized to zero bits. The constructors, if any, are ignored.
If T is a union type, the first non-static named data member is zero-initialized and all padding is initialized to zero bits.
If T is array type, each element is zero-initialized
If T is reference type, nothing is done.
gray_code
是非 union 类类型。因此,它的所有非静态数据成员都应该被初始化,这意味着 value
是零初始化的。 value
满足std::is_unsigned
,因此是一个标量类型,这意味着它应该用“整数常量零隐式转换为T”来初始化。
所以,如果我没看错的话,在上面的函数 foo
中,bar.value
应该总是用 0
初始化并且它永远不应该用垃圾初始化,对吗?
注意:我编译代码时使用的编译器是 MinGW_w4 GCC 4.9.1(POSIX 线程和 dwarf 异常),以防万一。虽然我有时会在我的计算机上得到垃圾,但我从来没有设法通过在线编译器得到零。
更新:似乎是一个 GCC 错误,错误是我的,而不是我的编译器的错误。实际上,在写这个问题时,为了简单起见,我假设
class foo {
foo() = default;
};
和
class foo {
foo();
};
foo::foo() = default;
是等价的。他们不是。这是来自 C++14 标准的引用,第 [dcl.fct.def.default] 部分:
A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration.
换句话说,当我得到垃圾值时,我的默认默认构造函数确实是用户提供的,因为它没有在第一次声明时显式错误。因此,发生的不是零初始化而是默认初始化。再次感谢@Columbo 指出真正的问题。
最佳答案
So, if I read correctly all of that, in the function
foo
above,bar.value
should always be initialized with0
and it should never be initialized with garbage, am I right?
是的。您的对象是直接列表初始化的。 C++14 的* [dcl.init.list]/3 指定
List-initialization of an object or reference of type
T
is defined as follows:
[… Inapplicable bullet points…]
Otherwise, if
T
is an aggregate, aggregate initialization is performed (8.5.1).Otherwise, if the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized.
[…]
您的类不是聚合,因为它具有用户提供的构造函数,但它确实具有默认构造函数。 [dcl.init]/7:
To value-initialize an object of type
T
means:
if
T
is a (possibly cv-qualified) class type (Clause 9) with either no default constructor (12.1) or a default constructor that is user-provided or deleted, then the object is default-initialized;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;
[dcl.fct.def.default]/4:
A special member function is user-provided if it is user-declared and not explicitly defaulted […] on its first declaration.
所以你的构造函数不是用户提供的,因此对象是零初始化的。 (构造函数没有被调用,因为它微不足道)
最后,如果不清楚,零初始化 T
类型的对象或引用意味着:
if
T
is a scalar type (3.9), the object is initialized to the value obtained by converting the integer literal0
(zero) toT
;if
T
is a (possibly cv-qualified) non-union class type, each non-static data member and each base-class subobject is zero-initialized and padding is initialized to zero bits;[…]
因此要么
你的编译器有问题
…或者您的代码在其他某个时间点触发了未定义的行为。
* 在 C++11 中答案仍然是肯定的,尽管引用的部分不等价。
关于c++ - 值初始化 : default initialization or zero initialization?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26699720/