std::erase_if (std::map)
From cppreference.com
Defined in header <map>
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(since C++20) | ||
Erases all elements that satisfy the predicate pred from c.
Equivalent to
auto old_size = c.size(); for (auto first = c.begin(), last = c.end(); first != last;) { if (pred(*first)) first = c.erase(first); else ++first; } return old_size - c.size();
Contents |
Parameters
c | - | container from which to erase |
pred | - | predicate that returns true if the element should be erased |
Return value
The number of erased elements.
Complexity
Linear.
Example
Run this code
#include <iostream> #include <map> void println(auto rem, auto const& container) { std::cout << rem << '{'; for (char sep[]{0, ' ', 0}; const auto& [key, value] : container) std::cout << sep << '{' << key << ", " << value << '}', *sep = ','; std::cout << "}\n"; } int main() { std::map<int, char> data { {1, 'a'}, {2, 'b'}, {3, 'c'}, {4, 'd'}, {5, 'e'}, {4, 'f'}, {5, 'g'}, {5, 'g'}, }; println("Original:\n", data); const auto count = std::erase_if(data, [](const auto& item) { auto const& [key, value] = item; return (key & 1) == 1; }); println("Erase items with odd keys:\n", data); std::cout << count << " items removed.\n"; }
Output:
Original: {{1, a}, {2, b}, {3, c}, {4, d}, {5, e}} Erase items with odd keys: {{2, b}, {4, d}} 3 items removed.
See also
removes elements satisfying specific criteria (function template) | |
(C++20)(C++20) |
removes elements satisfying specific criteria (niebloid) |