如果在std :: map中操作不存在的键/值对，会发生什么？

Question

Say we have a std::map, declared as:

std::map<int, int> m;

Then, we immediately do this:

m[0]++;

What happens? According to my little experiment, m[0] goes to '1', as if there had been a '0' stored for key '0'.

But what if the map contains, say, custom types?

std::map<E, T> m;
m[e].function();

Would this work?

Actually I was wondering how C++ standard specifies this:)

Answer 1

the overloaded operator[] for std::map actually creates the element passed as the argument, so by doing m[0]++ you insert an int and do ++ on it. Nothing wrong with it being 1

A call to this function is equivalent to:
(*((this->insert(make_pair(k,mapped_type()))).first)).second

Answer 2

The std::map will introduce an "empty" element when you use the operator[] to access an entry.

If you don't want there to be an empty element, then you need to use std::map::find() to check if the element is there, and if it's not, don't use it.

Of course, as long as your type T is fine to have function() called on it, nothing bad will happen from m[e].function() . If T needs to be initialized with something useful before function() can be called, you'll have to make sure this doesn't happen.