std::for_each - cppreference.com (original) (raw)

Applies the given unary function object f to the result of dereferencing every iterator in the range [first, last). If f returns a result, the result is ignored.

1. f is applied in order starting from first.

2. f might not be applied in order. The algorithm is executed according to policy.

This overload participates in overload resolution only if all following conditions are satisfied:

If the iterator type (InputIt/ForwardIt) is mutable, f may modify the elements of the range through the dereferenced iterator.

Unlike the rest of the parallel algorithms, for_each is not allowed to make copies of the elements in the sequence even if they are TriviallyCopyable.

Contents

• 1 Parameters
• 2 Return value
• 3 Complexity
• 4 Exceptions
• 5 Possible implementation
• 6 Notes
• 7 Example
• 8 Defect reports
• 9 See also

[edit] Parameters

[edit] Return value

1. f

2. (none)

[edit] Complexity

Exactly std::distance(first, last) applications of f.

[edit] Exceptions

The overload with a template parameter named ExecutionPolicy reports errors as follows:

• If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the standard policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
• If the algorithm fails to allocate memory, std::bad_alloc is thrown.

[edit] Possible implementation

See also the implementations in libstdc++, libc++ and MSVC stdlib.

template<class InputIt, class UnaryFunc> constexpr UnaryFunc for_each(InputIt first, InputIt last, UnaryFunc f) { for (; first != last; ++first) f(*first);   return f; // implicit move since C++11 }

[edit] Notes

For overload (1), f can be a stateful function object. The return value can be considered as the final state of the batch operation.

For overload (2), multiple copies of f may be created to perform parallel invocation. No value is returned because parallelization often does not permit efficient state accumulation.

[edit] Example

The following example uses a lambda-expression to increment all of the elements of a vector and then uses an overloaded operator() in a function object (i.k.a., "functor") to compute their sum. Note that to compute the sum, it is recommended to use the dedicated algorithm std::accumulate.

#include #include #include   int main() { std::vector v{3, -4, 2, -8, 15, 267};   auto print = [](const int& n) { std::cout << n << ' '; };   std::cout << "before:\t"; std::for_each(v.cbegin(), v.cend(), print); std::cout << '\n';   // increment elements in-place std::for_each(v.begin(), v.end(), [](int &n) { n++; });   std::cout << "after:\t"; std::for_each(v.cbegin(), v.cend(), print); std::cout << '\n';   struct Sum { void operator()(int n) { sum += n; } int sum {0}; };   // invoke Sum::operator() for each element Sum s = std::for_each(v.cbegin(), v.cend(), Sum());
std::cout << "sum:\t" << s.sum << '\n'; }

Output:

before: 3 -4 2 -8 15 267 after: 4 -3 3 -7 16 268 sum: 281

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[edit] See also