英文:
Visual Studio 2019 compiler error C2397 and warning C4838: narrowing conversion diagnostic
问题
在Visual Studio 2019中，以下代码会产生编译器错误C2397：
conversion from 'T' to 'size_t' requires a narrowing conversion.
与此相反，下面的代码会产生警告：
(level 1) C4838: "conversion from 'T' to 'size_t' requires a narrowing conversion".
我觉得这种行为不一致。我理解存在一个缩窄转换，C++标准要求生成诊断消息，但是第一个示例和第二个示例之间有何不同，使得第一个是错误，而第二个只是警告？
英文:
In Visual Studio 2019 the following code produces the compiler error C2397:
conversion from 'T' to 'size_t' requires a narrowing conversion.
struct Narrow
{
template <typename T>
Narrow(T t)
: szt{ t }
{ }
std::size_t szt[1];
};
Narrow narrow{ 3 };
In contrast, the code below produces the warning
(level 1) C4838: "conversion from 'T' to 'size_t' requires a narrowing conversion".
struct Narrow
{
template <typename T>
Narrow(T t)
{
std::size_t arr[] { t };
}
};
Narrow narrow{ 3 };
I find this behavior inconsistent. I understand that there is a narrowing conversion and C++ standard requires a diagnostic message, but what is so different between these examples such that the first one is an error, whereas the second is just a warning?
答案1
得分: 1
这是为了向后兼容而进行的不同行为。
T t{v} 在 C++11 之前是无效的代码。
T t = {v} 在 C++11 之前是有效的代码，并且可能会产生关于缩小转换的警告。
T t{v} 从 C++11 开始是有效的代码，可能会产生缩小转换的错误。这是符合标准 C++11 的新行为。它不会破坏旧代码，因为这是一种新的初始化语法。
关于T t = {v} 中的缩小转换警告是为了向后兼容而保留的。否则，如果将它与 C++11 编译器一起编译，它会破坏先前成功编译的 C++98 代码。
英文:
This is a different behavior for backward compatibilty.
T t{v} is an invalid code until C++11.
T t = {v} is a valid code until C++11 and can produce the warning about a narrowing conversation.
T t{v} is a valid code since C++11 and can produce an error of a narrowing conversion. It's a new behavior conforming to the standard C++11. It can't break an old code, because this a new initialization syntax.
The warning about a narrowing conversation in T t = {v} is kept for backward compatibilty. Otherwise it would break previously compiled successfully C++98 code if it's compiled with a C++11 compiler.
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