为什么这个复制初始化(有两个隐式转换)在 GCC/Clang 中不起作用?

Why this copy-initialization (with two implicit conversions) does not work in GCC/Clang?

请考虑以下代码:

#include <string.h>

class cstring {
public:
    cstring(const char* str) : begin_(str), end_(str ? str + strlen(str) : 0)
    {
    }

    size_t size() const
    {
        return (size_t)(end_ - begin_);
    }

    const char* data() const
    {
        return begin_;
    }

private:
    const char* begin_;
    const char* end_;
};

class Name {
public:
    Name(cstring str)
    {
    }
};

int main()
{
    cstring str = "Hello World"; // OK

    Name name = str; // OK

    Name name2 = "James"; // error
}

Clang 会输出错误(GCC 有类似的输出):

Source.cpp:37:10: error: no viable conversion from 'const char [6]' to 'Name'
    Name name2 = "James"; // error
         ^       ~~~~~~~
Source.cpp:24:7: note: candidate constructor (the implicit copy constructor) not viable: no known conversion from
      'const char [6]' to 'const Name &' for 1st argument
class Name {
      ^
Source.cpp:26:5: note: candidate constructor not viable: no known conversion from 'const char [6]' to 'cstring' for 1st
      argument
    Name(cstring str)
    ^
1 error generated.

上述代码中,由于构造函数不显式,const char*可以隐式转换为cstringcstring可以隐式转换为Name还有;因此存在从 const char*Name 的转换序列。那为什么 Clang/GCC 不允许这样的复制初始化(msvc 允许这种情况)?

gcc和clang是正确的,implicit conversion sequence中只允许一个用户自定义转换。

Implicit conversion sequence consists of the following, in this order:

  1. zero or one standard conversion sequence;

  2. zero or one user-defined conversion;

  3. zero or one standard conversion sequence.

您可以将 copy-initialization to direct-initialization 更改为 Name name2 ("James");,其中只需要一个用户定义的转换(从 const char*cstring),转换后的 cstring 作为参数传递给 Name 的构造函数以构造 name2.

In addition, the implicit conversion in copy-initialization must produce T directly from the initializer, while, e.g. direct-initialization expects an implicit conversion from the initializer to an argument of T's constructor.