std::map operator[] and automatically created new objects

I'm a little bit sca开发者_开发知识库red about something like this:

std::map<DWORD, DWORD> tmap;
  tmap[0]+=1;
  tmap[0]+=1;
  tmap[0]+=1;

Since DWORD's are not automatically initialized, I'm always afraid of tmap[0] being a random number that is incremented. How does the map know hot to initialize a DWORD if the runtime does not know how to do it?

Is it guaranteed, that the result is always tmap[0] == 3?

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Yes. When a new value is inserted into a map by operator[] it is value-initialized and for built-in numeric types (DWORD is a typedef for built-in type) this means zero.

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The new object, when inserted into the map by [] operator, is value-initialized. It is ensured by the map implementation, i.e. it is done "automatically" in that sense. For objects of type DWORD (assuming it is a scalar type), value-initialization means zero-initialization.

By definition given in 23.3.1.2, operator [] is a shorthand for

(*((insert(make_pair(x, T()))).first)).second

The T() bit is the new object, which will turn into DWORD() in your case. DWORD() is guaranteed to be zero.

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Yes. If the key you passed to operator[] doesnot exist, then map will default construct the object and inserts it. In your case it will do DWORD() which will yield a value 0.