C++ SFINAE在引用类型和转换为引用的参数之间选择重载？

Question

考虑一下：

struct BigData {...};

// Let's get a BigData by reference, but use it as a value.
// For example, we may want to make a copy of the object, but we'd
// like to avoid the pass-by-value overhead at the call site.
template <typename T, *some template magic?*>
void processData(T& t) {
    printf("Your BigData was converted to a reference argument.\n");
    ...
}

// Now, we want an overload that will know that there wasn't a
// conversion to reference and will treat it like a reference.
// Perhaps we are adding this BigData to a list of references.
template <typename T, *some template magic?*>
void processData(T& t) {
    printf("You gave me a BigData reference.\n");
    ...
}

int main() {
    BigData data;
    BigData& ref = data;

    processData(data); // "Your BigData was converted to a reference argument."
    processData(ref); // "You gave me a BigData reference."

    return 0;
}

简而言之，我的目标是让重载区分引用绑定来自何处-要么是类型的值，要么是(已经)引用类型。我曾尝试将std::enable_if和family与不引用和不引用T的重载结合使用，但我找不到一种方法来实现这一点。任何帮助都是非常感谢的！

Answer 1

我认为你没有理解引用绑定是如何工作的。您不仅可以将一个引用绑定到另一个引用，还可以绑定一个值类型的值(即不限定引用)

因此，如果希望通过引用传递BigData对象，只需执行以下操作

template <typename T>
void processData(T& t) {
    cout << "Your BigData was passed as a reference argument" << endl;
    // ...
}

int main() {
    BigData data;
    BigData& ref = data;

    processData(data); 
    processData(ref);

    return 0;
}

在这里，两个processData调用都将通过引用传递BigData对象(即在函数调用期间没有副本)。

当传递给函数的对象是独立于正常情况的引用时，您不需要处理这种情况。这里不需要std::enable_if。

此外，没有办法区分这两种情况，使用data或ref作为函数的参数，因为它们都是左值。如果要检查表达式的decltype是否为引用，则必须执行类似以下操作

#include <iostream>
#include <type_traits>

using std::cout;
using std::endl;

template <typename T, std::enable_if_t<std::is_reference<T>::value>* = nullptr>
void processData(std::add_lvalue_reference_t<T>) {
    cout << "You gave me a reference argument." << endl;
}

template <typename T, std::enable_if_t<!std::is_reference<T>::value>* = nullptr>
void processData(std::add_lvalue_reference_t<T>) {
    cout << "Your argument was converted to a reference." << endl;
}

int main() {
    auto integer_value = 1;
    const auto& integer_ref = 2;
    processData<decltype(integer_value)>(integer_value);
    processData<decltype(integer_ref)>(integer_ref);

    return 0;
}