CODEWITHSUNDEEP
c++stl
kevinshms
kevinshms

Reputation: 41

STL std::map operator[] when on the right side of assignment

I am aware that the map dereference operator[] is not const and can modify the map. However, I was wondering what happens when the operator is on the right side. For instance:

std::map<int, int> a;
int b = a[0];

Since a will not have the 0 entry will this create a new entry in the map with the key 0?

Upvotes: 2

Views: 646

Answers (2)

Jun Ge
Jun Ge

Reputation: 420

Yes, map will create a new entry with key 0 and value 0 in your case.

Below is the description of operation [] in http://www.cplusplus.com/reference/map/map/operator[]/

mapped_type& operator[] (const key_type& k);

If k matches the key of an element in the container, the function returns a reference to its mapped value.If k does not match the key of any element in the container, the function inserts a new element with that key and returns a reference to its mapped value.

And sometimes it will introduce upexpected behavior if people forget the operator [] will insert a new key-value pair when it doesn't exist in the map. For example, below is a usural mistake.

std::map<int,int> my_map;

//check if 100 is in my_map
if ( my_map[100] ) {
    //some code
}

I just want to check whether 100 is in my_map or not, but above code will insert a new key-value in my_map. So in this case, I should use find instead of operator [] if I just want to check 100.

std::map<int,int> my_map;

//check if 100 is in my_map
if ( my_map.find(100) != my_map.end() ) {
    //some code
}

Upvotes: 1

Weak to Enuma Elish
Weak to Enuma Elish

Reputation: 4637

Regarding operator[]:

Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.

From here, emphasis mine.

This is easy to test, too.

std::map<int, int> a;
std::cout << a.size(); //0
int b = a[0];
std::cout << a.size(); //1

Live example: http://ideone.com/z7PIRx

Upvotes: 6

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