返回函数局部变量作为参考 [英] Returning function-local variable as reference

问题描述

我发现这个源代码：

inline GUID& WString2Guid(wstring src)
{
    static GUID result;
    HRESULT hr = ::CLSIDFromString(W2OLE(const_cast<WCHAR*>(src.c_str())), &result);
    if (FAILED(hr)) {
        //ERROR: The string '%s' is not formatted as a GUID!
        throw(E_INVALIDARG);
    }
    return result;
}

在这里返回引用的用法是什么？调用代码不能得到一个引用，因为变量会在那时离开它的范围。

What's the use of returning a reference here? The calling code cannot get a reference anyway because the variable would have left its scope by then. So does this little & sign make any difference?

为了澄清/扩展问题：在同一个例子中程序，该函数被调用为

To clarify/extend the question: In the same example program, the function is called as

GUID guid = WString2Guid(id); // way 1

如果我想使用引用，

GUID& guid = WString2Guid(id); // way 2

？

另一个问题;为什么 :: 范围操作符之前调用 CLSIDFromString 如果有另一个局部函数声明有相同的名字，这将是有意义的，不是吗？

And another question; why is the CLSIDFromString function called with the :: scope operator before? This would only make any sense if there was another local function declared with the same name, wouldn't it?

推荐答案

result 是一个 static 局部变量，因此即使在函数退出后它也会存在。不要将其与非静态局部变量混淆。

No. result is a static local variable, so it will exist even after the function exit. Don't confuse this with non-static local variable.

:: ：CLSIDFromString 告诉编译器从全局命名空间中选择 CLSIDFromString ，如果存在 CLSIDFromString 在其他命名空间中定义，在调用站点可见。

:: in ::CLSIDFromString tells the compiler to choose CLSIDFromString from the global namespace, in case if there are many definition of CLSIDFromString defined in other namespace(s), visible at the call-site.

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