C++ template

Template Argument Deduction For Function Calls

The ultimate source of truth is temp.deduct.call. Let’s take a few examples.

Example 1: cv-qualifier and reference type

template <typename T> void foo(const T& x);
int x = 42;
const int y = 42;
foo(x); // A = int. P = const int&
foo(y); // A = const int. P = const int&

Both int x and const int y are deduced as int. This is because p2.3:

If A is a cv-qualified type, the top-level cv-qualifiers of A’s type are ignored for type deduction.

So in both cases, A is adjusted to int. The corresponding LLVM implementation is here. Also, p3 says

If P is a cv-qualified type, the top-level cv-qualifiers of P’s type are ignored for type deduction. If P is a reference type, the type referred to by P is used for type deduction.

Here P is const T&, i.e., a reference to const T. Note, const is not at top-level. So P is adjusted to const T. The corresponding LLVM implementation is here. Now, the task is to solve equation int = const T. It seems impossible because the extra const modifier, but then how compiler deduces T = int? This is because p4.1

If the original P is a reference type, the deduced A (i.e., the type referred to by the reference) can be more cv-qualified than the transformed A.

Example 2: forward reference

The examples given in p3 are clear, and no need to repeat here. First, it gives a precise definition of forward reference.

A forwarding reference is an rvalue reference to a cv-unqualified template parameter that does not represent a template parameter of a class template.

For a template argument to be qualified as a forward reference, it must satisfies 3 criteria. First, it is a rvalue reference T&&. Second, it should not have cv-qualifier. const T&& is not a forward reference. Third, the template parameter should not belong to the class template. So basically, a function using forward reference can only take the form template <class T> foo(T&& t) or template <typename... Args> foo(Args&&... args).

Also, p3 states the special treatment of forward reference in type deduction.

If P is a forwarding reference and the argument is an lvalue, the type “lvalue reference to A” is used in place of A for type deduction.

This means that for template <class T> foo(T&& t); int x = 5; foo(x);, we solve T&& = int&, so we get T = int&. You can verify that with this special rule, no matter what the input argument is, the resulting type for T is either lvalue reference or rvalue reference. That is why it is initially called universal referencee. Later on, the committee standardized the name to forward reference to emphasize its intentional usage: forward reference is used to forward the argument to the downstream callees.It is not supposed to be used directly in the current function. You can see more rigid reasoning in N4164.

Perfect Forwarding

Before understanding why we need perfect forwarding, let’s take a look at below code.

int&& r = 10;  // r is an rvalue reference to the temporary value 10
r = 20;         // r is now an lvalue, since we can assign to it

We definitely know that r is a rvalue reference. But rvalue references themselves are lvalues. r has a name, shows up on the left side and persists in memory, so it is lvalue. To extend this fact, given a function void foo(int&& x);, x has lvalue semantics inside the function body no matter what the input argument is. This is the exact problem for below code

#include <iostream>

void bar(int& x) { std::cout << "Lvalue reference\n"; }
void bar(int&& x) { std::cout << "Rvalue reference\n"; }
template <typename T> void foo(T&& arg) { bar(arg); }

int main() {
    int x = 42;
    foo(x);   // Expected: Lvalue reference -> Works fine
    foo(42);  // Expected: Rvalue reference -> WRONG! Calls Lvalue version
}

We need to find a way to perfectly pass forward reference arguments to the callee. Perfect forwarding allows a template function that accepts a set of arguments to forward these arguments to another function whilst retaining the lvalue or rvalue nature of the original function arguments. It goes like this

template<class T>
void foo(T&& a) {
  bar(std::forward<T>(a));
}

Implementation of std::forward is simple. It is just a static_cast.

Tricks

Use class... Args to mimic Python *args and **kwargs. See example.