Suppose I have a map
std::map<int, double> foo;
Is the behaviour on my writing foo[2] += 3.0; defined? That is, are any implicitly added map elements automatically initialised (hopefully to 0.0) in my case?
If not, am I introducing a truck-load of undefined behaviour? If so, could I do something funky with an allocator to enforce initialisation to 0.0?
Yes, it will be value-initialized (as 0.0 in your case). According to cppreference:
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.
If an insertion is performed, the mapped value is value-initialized (default-constructed for class types, zero-initialized otherwise) and a reference to it is returned.
new T(). "Scalars and POD types with dynamic storage duration were considered to be not initialized (since C++11, this situation was reclassified as a form of default initialization). " That's why I'm curious if C++03 really is the same as C++11 as regards this question, perhaps a similar change (similar to that quote) has been introduced for map elements. But I suppose if there was a difference, the page you linked would say so.
N3337 [map.access]/1Effects: If there is no key equivalent toxin the map, insertsvalue_type(x,T())into the map.
T() is value-initialization, which is the case of built-in types causes zero-initialization. As such, foo[2] will insert a zero-initialized double into your map, so your code is well-defined.
Yes, they are automatically value initialized when using operator[] on a non existing key. Specifically in the standard it's described at §23.4.4.3/1 (when talking about operator[]):
Effects: If there is no key equivalent to x in the map, inserts
value_type(x, T())into the map.
For most numeric types, including double, the expression T() yields a value-initialized element of that type, therefore yielding 0.0 in your case.
double doesn't really have a default constructor.
double.