std::map, insert() issue

Is there a specific reason (and if yes, what) that std::map insert() does not have the Obvious (to me at least) override defined as:

Code:

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some_return_type insert( const key_type& key, const mapped_type& mapped);

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I'm not sure what the ideal return type should be in that case, I guess it could be the same as the one for insert with const value_type& (so returning a pair<iterator, bool>).

using the [] operator isn't always possible since that assumes/requires that there is a default constructor for mapped_type. (with an assortment of potential performance issues)

compare the Obvious (?)

Code:

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foobarmap.insert(foo, bar);

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vs. the somewhat more awkward

Code:

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foobarmap.insert( std::pair<fooType,barType>(foo, bar));

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this can be particularly annoying if the types are complex constructs of their own (even if you typedef it)

Quote:

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Originally Posted by OReubens

Is there a specific reason (and if yes, what) that std::map insert() does not have the Obvious (to me at least) override defined as:

Code:

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some_return_type insert( const key_type& key, const mapped_type& mapped);

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Maybe because the key type and mapped types could go through a silent conversion, given you the other two argument constructor being called (which could lead to hard to find bugs even though the code would compile).

http://www.cplusplus.com/reference/map/map/insert/

For example:

Code:

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std::map<char *, char*>

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If your proposal of the two argument constructor were used, both char* arguments would silently convert into the (iterator, iterator) 2-arg constructor, and not the desired pair<char*, char*> that would be expected.

Regards,

Paul McKenzie

Quote:

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Originally Posted by OReubens

this can be particularly annoying if the types are complex constructs of their own (even if you typedef it)

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There is always std::make_pair(...) to prevent some repetition.

Quote:

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Originally Posted by Paul McKenzie

For example:

Code:

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std::map<char *, char*>

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If your proposal of the two argument constructor were used, both char* arguments would silently convert into the (iterator, iterator) 2-arg constructor, and not the desired pair<char*, char*> that would be expected.

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I think in this case overload resolution would pick the non-template overload, so it would pick the (key_type, mapped_type) overload. However, with a std::map<std::string, std::string> and calling insert("key", "value"), I think the template (Iter, Iter) overload would be picked.