未调用复制构造函数进行复制初始化或优化？
如果复制构造函数设为私有，那么在
情况1：没有错误的情况下，编译器不关心复制构造函数是否在类中定义。
情况 2： 错误，复制构造函数是私有的，当它成为公共时，它会被忽略。
它是否直接优化副本而没有注意到构造函数是否被设为私有？
#include <string>
using std::string;
class T
{
string s;
T(const T &obj):s(obj.s){}
public:
T(const string &str):s(str){}
};
int main()
{
T a = ("Copy Initialization"); //Case: 1
T b = T("Copy Initialization"); //Case: 2
}
If copy constructor is made private then in
Case 1: No error, the compiler doesn't care if the copy constructor was defined in class.
Case 2: Error, copy constructor is private and when it is made public, it is elided.
Does it directly optimizes the copy without being noticing that if the constructor was made private?
#include <string>
using std::string;
class T
{
string s;
T(const T &obj):s(obj.s){}
public:
T(const string &str):s(str){}
};
int main()
{
T a = ("Copy Initialization"); //Case: 1
T b = T("Copy Initialization"); //Case: 2
}
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评论（2）
案例 2 在 N3225 中低于 12.8/31：
仅仅因为复制因子被省略并不意味着它没有被 ODR 使用。 3.2/2：
当然要注意，MSVC 不完全兼容 C++0x，因为 (a) C++0x 还不是标准，并且尚未最终确定； (b) MSVC 还没有实现最新的所有内容。但这个东西与 C++03 相比并没有实质性改变，所以我相当有信心这个解释仍然成立。
情况 1 也会属于这种情况，除了在我检查过的两个 C++03 编译器上它没有达到那么远，因为不可能从字符串文字到 T 的转换。我懒得去检查是否C++0x 中允许任何额外的转换序列，任何地方都可能有一个新子句:-)
对我来说，为什么 MSVC 允许情况 1 仍然是个谜，即使有一个公共复制构造函数。它允许在严格的 C++03 模式下使用吗？
Case 2 comes under 12.8/31 in N3225:
Just because the copy ctor is elided doesn't mean it isn't odr-used. 3.2/2:
Beware of course that MSVC is not fully C++0x-compliant, because (a) C++0x isn't a standard yet, and isn't finalized; and (b) MSVC hasn't implemented everything up to date anyway. But this stuff isn't substantially changed from C++03, so I'm fairly confident the explanation still holds.
Case 1 would come under this too, except that on the two C++03 compilers I've checked it doesn't get that far because there's no possible conversion from a string literal to T. I can't be bothered to check whether there are any additional conversion sequences allowed in C++0x, there could be a new clause anywhere :-)
It's still a mystery to me why MSVC allows case 1 ever, even with a public copy ctor. Does it allow it in strict C++03 mode?
T a = ("Copy Initialization");应该给出错误，因为没有合适的从"const char [20]"转换为"T"的构造函数您的意思是
T a = std::string("Copy Initialization");< /代码>？不，不能。编译器通常在代码优化阶段之前执行语法和语义分析。
T a = ("Copy Initialization");should give an error because there's no suitable constructor to convert from"const char [20]"to"T"Did you mean
T a = std::string("Copy Initialization");?No it can't. Compilers usually perform syntactic and semantic analysis prior to the code optimization phase.