Fold Expressions

Remarks

Fold Expressions are supported for the following operators

            
+-*/%&|<<>>
+=-=*=/=%=\ˆ=&=|=<<=>>==
==!=<><=>=&&||,.*->*

When folding over an empty sequence, a fold expression is ill-formed, except for the following three operators:

OperatorValue when parameter pack is empty
&&true
||false
,void()

Binary Folds

Binary folds are basically unary folds, with an extra argument.

There are 2 kinds of binary folds:

  • Binary Left Fold - (value op ... op pack) - Expands as follows:

    (((Value op Pack1) op Pack2) op ...) op PackN
    
  • Binary Right Fold (pack op ... op value) - Expands as follows:

    Pack1 op (... op (Pack(N-1) op (PackN op Value)))
    

Here is an example:

template<typename... Ts>
int removeFrom(int num, Ts... args)
{
    return (num - ... - args); //Binary left fold
    // Note that a binary right fold cannot be used
    // due to the lack of associativity of operator-
}

int result = removeFrom(1000, 5, 10, 15); //'result' is 1000 - 5 - 10 - 15 = 970

Folding over a comma

It is a common operation to need to perform a particular function over each element in a parameter pack. With C++11, the best we can do is:

template <class... Ts>
void print_all(std::ostream& os, Ts const&... args) {
    using expander = int[];
    (void)expander{0,
        (void(os << args), 0)...
    };
}

But with a fold expression, the above simplifies nicely to:

template <class... Ts>
void print_all(std::ostream& os, Ts const&... args) {
    (void(os << args), ...);
}

No cryptic boilerplate required.

Unary Folds

Unary folds are used to fold parameter packs over a specific operator. There are 2 kinds of unary folds:

  • Unary Left Fold (... op pack) which expands as follows:

    ((Pack1 op Pack2) op ...) op PackN
    
  • Unary Right Fold (pack op ...) which expands as follows:

    Pack1 op (... (Pack(N-1) op PackN)) 
    

Here is an example

template<typename... Ts>
int sum(Ts... args)
{
    return (... + args); //Unary left fold
    //return (args + ...); //Unary right fold

    // The two are equivalent if the operator is associative.
    // For +, ((1+2)+3) (left fold) == (1+(2+3)) (right fold)
    // For -, ((1-2)-3) (left fold) != (1-(2-3)) (right fold)
}

int result = sum(1, 2, 3); //  6


2016-07-22
2016-10-28
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