Fold expression

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Quick fact

• Since C++17
• Reduces (folds) a parameter pack over a binary operator.
  • e.g. They allow you to "fold" the operation over the elements of the pack, combining them into a single result.
  • "A fold expression is an instruction for the compiler to repeat the application of an operator over a variadic template pack."
• Fold expressions provide a concise and powerful way to work with parameter packs, reducing the need for recursion or explicit loops when operating on variadic template arguments.

Syntax

• The position of ... specifies left or right associativity, but doesn't change the order of arguments.
• Four syntax (Note that the opening and closing parentheses are a required part of the fold expression.):

( pack op ... )	// unary right fold
( ... op pack )	// unary left fold
( pack op ... op init )	// binary right fold
( init op ... op pack )	// binary left fold

• 1. Unary right fold (E op ...) becomes (E1 op (... op (EN-1 op EN)))
• 2. Unary left fold (... op E) becomes (((E1 op E2) op ...) op EN)
• 3. Binary right fold (E op ... op I) becomes (E1 op (... op (EN−1 op (EN op I))))
• 4. Binary left fold (I op ... op E) becomes ((((I op E1) op E2) op ...) op EN)

For example:

// Example of 2: ( ... op pack )
template<typename... Args>
bool all(Args... args) { return (... && args); }

bool b = all(true, true, true, false);
 // within all(), the unary left fold expands as
 //  return ((true && true) && true) && false;
 // b is false

// Example of 4: ( init op ... op pack )
template<typename ...Args>
void printer(Args&&... args)
{
    (std::cout << ... << args) << '\n';
}

// My aha moment!! What if I do...
template<typename ...Args>
void printer2(Args&&... args)
{
    std::cout << (... << args) << '\n';
}
printer2(1, 2, 3, 4); // result: 512
// why? This becomes form 2: ( ... op pack )
// so it basically becomes...
// std::cout << (((1 << 2) << 3) << 4) << '\n';
// e.g. 4 * 8 * 16 = 512!!!!