默认构造函数，POD的初始化和C ++ 11中的隐式类型转换 [英] Default constructors, initialization of POD and implicit type conversions in C++11
问题描述
我刚看过Chandler在Going Native 2012上的Clang演讲。他提供了以下代码：
I have just watched Chandler's presentation on Clang at Going Native 2012. He presents the following code:
#include <iostream>
struct S{ int n; };
struct X{ X(int) {}; };
void f( void* )
{
std::cerr << "Pointer!\n";
}
void f( X )
{
std::cerr << "X!\n";
}
int main()
{
f(S().n);
}
Chandler声明这会调用 f 用于c ++ 11和 f（X）用于c ++ 03。
他还说，原因是S（）.n默认初始化为0，使它成为一个 nullptr 常量。
Chandler states that this calls f(void*) for c++11 and f(X) for c++03.
He also states that the reason is that S().n is default initialised to 0, making it a nullptr constant.
首先我假定成员变量n的零初始化是依赖于编译器实现的，并且不被标准保证（或者使用c ++ 11进行这种更改）？ Chandler提示这是由于支持常量表达式，但我仍然不能完全按照他的推理。
Firstly am I right in assuming that the zero initialization of member variable n is compiler implementation dependent and NOT guaranteed by the standard (or did this change with c++11)? Chandler hints this is due to support for constant expressions but I still cannot fully follow his reasoning.
其次为什么 f（X）用C ++ 03而不是c ++ 11调用？我会假设 f（void *）将会启动，而不管 S（）的值。n over隐式转换为 X
Secondly why would f(X) be called with C++03 and not c++11? I would of assumed that f(void*) would kick in regardless of the value of S().n over an implicit conversion to X
对于Chandler的解释，看下面的链接，45分钟：
For Chandler's explanation see the following link, 45 minutes in:
推荐答案
我是正确的假设成员
变量n的零初始化是编译器实现依赖和不保证由
标准（或与c + + 11这样的更改）？
Firstly am I right in assuming that the zero initialization of member variable n is compiler implementation dependent and NOT guaranteed by the standard (or did this change with c++11)?
否， S（）表示在C ++ 03和C ++ 11中的值初始化。虽然我相信这个词在C ++ 11比C ++ 03更清楚。在这种情况下，值初始化转向零初始化。对比此默认初始化不为零：
No, S() means value initialize in both C++03 and C++11. Though I believe the wording for this is much clearer in C++11 than C++03. In this case, value-initialization forwards to zero-initialization. Contrast this with default-initialization which does not zero:
S s1; // default initialization
std::cout << s1.n << '\n'; // prints garbage, crank up optimizer to show
S s2 = S(); // value initialization
std::cout << s2.n << '\n'; // prints 0
其次，为什么要调用f C ++ 03而不是c ++ 11？我想
假设f（void *）将在不考虑S（）的值的情况下进入。n
对X的隐式转换
Secondly why would f(X) be called with C++03 and not c++11? I would of assumed that f(void*) would kick in regardless of the value of S().n over an implicit conversion to X
在C ++ 03中， int 永远不能成为空指针常量。一旦 0 键入为 int ，则将其分配给 int ，那么它永远是 int ，而不是空指针常量。
In C++03 an int can never become a null pointer constant. Once 0 is typed as an int, say by assigning it to an int, then it is forever an int, and not a null pointer constant.
+11， S（）。n 隐含地是 constexpr code>，并且值为 0 的 constexpr 表达式可以是空指针常量。
In C++11, S().n is implicitly a constexpr expression with value 0, and a constexpr expression with value 0 can be a null pointer constant.
最后，就我所知，这不是委员会的有意改变。如果你编写的代码取决于这种差异，你可能会被咬在未来如果/当委员会纠正自己。我会引导远离这个领域。使用它来赢得投注 - 而不是在生产代码。
Finally, this is not an intentional change on the part of the committee as far as I can tell. If you're writing code that depends on this difference, you may well get bitten in the future if/when the committee corrects itself. I would steer well clear of this area. Use it to win bar bets -- not in production code.
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