std::unordered_set::erase <cpp |container |unordered set iterator erase(const_iterator pos); (1)(since C++11) iterator erase(const_iterator first, const_iterator last); (2)(since C++11) size_type erase(constkey_type&key);
#include <unordered_set> #include <iostream> int main() { std::unordered_set<int> c = {1, 2, 3, 4, 5, 6, 7, 8, 9}; // 从 c 擦除所有奇数 for(auto it = c.begin(); it != c.end(); ) if(*it % 2 == 1) it = c.erase(it); else ++it; for(int n : c) std...
std::unordered_set<Key, Hash, KeyEqual, Alloc>::size_type erase_if(std::unordered_set<Key, Hash, KeyEqual, Alloc>&c, Pred pred); (since C++20) Erases all elements that satisfy the predicatepredfromc. Equivalent to autoold_size=c.size();for(autofirst=c.begin(), last=c.end();fir...
std::unordered_set (1) iterator erase(iterator pos); (since C++11) (until C++23) iterator erase(iterator pos) requires(!std::same_as<iterator, const_iterator>); (since C++23) iterator erase(const_iterator pos); (2)(since C++11) ...
voiderase_if(std::unordered_set<Key,Hash,KeyEqual,Alloc>&c, Pred pred); (1)(C++20 起) 从容器中擦除所有满足谓词pred的元素。等价于 for(autoi=c.begin(), last=c.end();i!=last;){if(pred(*i)){i=c.erase(i);}else{++i;}} ...
相反,在unordered_map上,只有那些实际上删除了特定元素的操作才会使引用无效,即使是rehash(可能是分配...
相反,在unordered_map上,只有那些实际上删除了特定元素的操作才会使引用无效,即使是rehash(可能是分配...
我有点担心.reserve()实际上阻止了以后的运行时内存分配(我并不真正理解wrt用于堆使用的解释),但特别是.erase(),我无法保证它在调用时不会请求堆进行动态去分配。 因此,问题是std::unordered_map::erase的指定堆交互(如果有的话)是什么,如果它确实执行释放,那么是否有某种技巧可以用来避免它们呢?
std::unordered_set<unsigned int> insert_microseconds=843727, query_microseconds=104578, erase_microseconds=299443, all_microseconds=1247748 测试代码如下: int main( int argc, char *argv[] ) { char chars_tmp[24]; std::string str_msg;
#include <algorithm> #include <iostream> #include <unordered_set> int main() { std::unordered_set<int> container{1, 2, 3}; auto print = [](const int& n) { std::cout << " " << n; }; std::cout << "Before clear:"; std::for_each(container.begin(), container.end(), prin...