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Disabling Template Argument Deduction in std::forward
Question:
Why is template argument deduction disabled in the implementation of std::forward in Visual Studio 2010?
Answer:
std::forward is a utility function used for perfect forwarding, where the type of the input argument is preserved in the output argument. However, if template argument deduction were enabled in std::forward, it could lead to incorrect behavior in certain scenarios.
Specifically, if the input argument is an lvalue or const lvalue (i.e., an object with a name), template argument deduction would deduce the parameter type to be an lvalue or const lvalue reference. In the perfect forwarding scenario, this is not what is desired, as the goal is to retain the exact type of the input argument.
To illustrate the potential issue:
<code class="cpp">template<typename T> void test(T&& obj) { // ... } int main() { int x; int& y = x; test(7); // input is an rvalue, so T is deduced as int&& test(x); // input is an lvalue, so T is deduced as int& test(y); // input is a named lvalue, so T is deduced as int& }</code>
In this example, if std::forward used template argument deduction, the type of the argument in test(x) and test(y) would be deduced incorrectly as int&, leading to unexpected behavior.
To prevent this issue, the implementation of std::forward uses the identity metafunction to explicitly specify the type parameter, effectively disabling template argument deduction. This ensures that the type of the input argument is preserved in the output argument regardless of whether it is an rvalue or an lvalue.
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