std :: unordered_map的以下命令之间是否有任何区别

Question

For inserting a key-value pair into an unordered map, eg, std::unordered_pair<int,int> map1 , can we do it in any of these two ways:

map1[2]=5;
map1.insert({2,5});

Is there any difference between using std_unordered_insert or operator[] ?

And if I want to find the mapped value for a given key, can I use either of the following:

mappedVal = map1.at(2);
mappedVal = map1[2];

Again, any difference between using std::unordered_map::at or operator[] ?

Answer 1

map1[2]=5;

If an entry with key 2 exists, set that entry's value to 5 . Otherwise, create a new entry with key 2 and value 5 .

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map1.insert({2,5});

If no entry with key 2 exists, create a new entry with key 2 and value 5 . Otherwise, do nothing .

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mappedVal = map1.at(2);

If an entry with key 2 exists, assign its value to mappedVal . Otherwise, throw an out_of_range exception.

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mappedVal = map1[2];

If an entry with key 2 exists, assign its value to mappedVal. Otherwise, create an entry for 2 using the default value and assign that default value to mappedVal.

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For lookups, I usually use unordered_map::find() rather than at() or operator [] () (unless I know that there is an entry for the given key).

Answer 2

In contrast to operator[] , at() will throw a std::out_of_range exception if the key doesn't exist. operator[] will create the key instead.