Question

我试图了解隐式类成员初始化如何对成员{POD结构，POD类和POD}起作用。在阅读了一下后，我预计它们会被初始化为默认值，但实际行为似乎有所不同 -

#include <iostream>

struct S1
{
    void* a;
    int b;
};

struct S2
{
    S2() { std::cout << "!"; }
    void* a;
    int b;
};

struct S3
{
    S3() : a(), b() { std::cout << "!"; }
    void* a;
    int b;
};

class C1
{
public:
    void* a;
    int b;
};

class C2
{
public:
    C2() { std::cout << "!"; }
    void* a;
    int b;
};

class C3
{
public:
    C3() : a(), b() { std::cout << "!"; }
    void* a;
    int b;
};


template <typename T>
class FOO1
{
public:
    T s;
    int a;
};

template <typename T>
class FOO2
{
public:
    FOO2() {}
    T s;
    int a;
};

template <typename T>
class FOO3
{
public:
    FOO3() : s(), a() {}
    T s;
    int a;
};

//#define SKIP_S1C1

template <typename T>
void moo()
{
#ifndef SKIP_S1C1
    T* f = new T();
    T foo = *f;
    std::cout << ":\ts = (" << foo.s.a << ", " << foo.s.b << ")\ta = " << foo.a << std::endl;
    delete f;
#else
    T foo;
    std::cout << ":\ts = (" << foo.s.a << ", " << foo.s.b << ")\ta = " << foo.a << std::endl;
#endif
}


int main()
{
#ifndef SKIP_S1C1
    moo<FOO1<S1> >();
#endif
    moo<FOO1<S2> >();
    moo<FOO1<S3> >();
#ifndef SKIP_S1C1
    moo<FOO1<C1> >();
#endif
    moo<FOO1<C2> >();
    moo<FOO1<C3> >();

std::cout << std::endl;

#ifndef SKIP_S1C1
    moo<FOO2<S1> >();
#endif
    moo<FOO2<S2> >();
    moo<FOO2<S3> >();
#ifndef SKIP_S1C1
    moo<FOO2<C1> >();
#endif
    moo<FOO2<C2> >();
    moo<FOO2<C3> >();

std::cout << std::endl;

#ifndef SKIP_S1C1
    moo<FOO3<S1> >();
#endif
    moo<FOO3<S2> >();
    moo<FOO3<S3> >();
#ifndef SKIP_S1C1
    moo<FOO3<C1> >();
#endif
    moo<FOO3<C2> >();
    moo<FOO3<C3> >();
}

明显的运行结果不足以说明POD是初始化为默认值0还是仅包含噪声。但无论如何，这里有一些结果：

在ubuntu上使用gcc 4.6.3构建并运行它#define SKIP_S1C1取消注释，我得

!:      s = (0x7ffffe557770, 4196620)   a = 1
!:      s = (0, 0)      a = 1
!:      s = (0, 0)      a = 1
!:      s = (0, 0)      a = 1

!:      s = (0x1, 6299744)      a = 6299744
!:      s = (0, 0)      a = 6299744
!:      s = (0, 0)      a = 6299744
!:      s = (0, 0)      a = 6299744

!:      s = (0x1, 6299744)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0

随着它的注释，我得到了

:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0

:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0

:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
:       s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0
!:      s = (0, 0)      a = 0

和VS2013一起评论，

:       s = (00000000, 0)       a = 0
!:      s = (CDCDCDCD, -842150451)      a = -842150451
!:      s = (00000000, 0)       a = -842150451
:       s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

:       s = (CDCDCDCD, -842150451)      a = -842150451
!:      s = (CDCDCDCD, -842150451)      a = -842150451
!:      s = (00000000, 0)       a = -842150451
:       s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

:       s = (00000000, 0)       a = 0
!:      s = (CDCDCDCD, -842150451)      a = 0
!:      s = (00000000, 0)       a = 0
:       s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

并取消注释，

!:      s = (CCCCCCCC, -858993460)      a = -858993460
!:      s = (00000000, 0)       a = -858993460
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

!:      s = (CCCCCCCC, -858993460)      a = -858993460
!:      s = (00000000, 0)       a = -858993460
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

!:      s = (CCCCCCCC, -858993460)      a = 0
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0
!:      s = (00000000, 0)       a = 0

我真的很想了解{POS struct，POD classes和POD}成员的隐式初始化时我应该期待什么以及何时是UB。任何帮助将不胜感激...... :)

Answer 1

构造函数很复杂，细节是技术性的，但这里是一个通用摘要*：

初始化有三种方法：

零初始化 - 细节是技术性的，但有效地将所有位设置为零。这绕过了构造函数
默认初始化 - 如果它有构造函数，则调用默认构造函数。否则，不会进行初始化。 从这些中读取的是您找到的UB。
Value Initialize - 如果它有构造函数，则调用默认构造函数。否则，它的所有（有效）位都设置为零。

他们在很多情况下被召唤：

static globals - Zero Initialized，然后Value Initialized。（非常奇怪）
locals - 默认初始化。
new T; - 默认初始化
new T(); - 值初始化
成员不在初始列表中 - 默认初始化
init list中的成员 - Value Initialize。

有关更多详细信息，请参阅C ++ 11草案中的§8.5和§12.6部分。他们很长很乏味。

另请注意，C的规则在技术上有惊人的不同，尽管效果对我来说是一样的。

*我的摘要在技术上并不准确，但对于大多数实际代码而言足够准确。例如，数组在技术上有特殊的规则，但它们非常直观，不值得一提。

**是的，它的“初始化”是“无初始化”，这使得其他段落“如果它被初始化”在技术上含糊不清，但应用常识。它没有初始化。

（POD struct，POD class和POD）成员的c ++隐式类成员初始化

1 个答案: