A recent talk about unordered_map in C++ made me realize that I should use unordered_map for most cases where I used map before, because of the efficiency of lookup ( amortized O(1) vs. O(log n) ). Most times I use a map, I use either int or std::string as the key type; hence, I've got no problems with the definition of the hash function. The more I thought about it, the more I came to realize that I can't find any reason of using a std::map over a std::unordered_map in the case of keys with simple types -- I took a look at the interfaces, and didn't find any significant differences that would impact my code.
因此，问题是:是否有任何真正的理由使用std::map而不是std::unordered_map在简单类型的情况下，如int和std::string?
我是从严格的编程角度提出这个问题的——我知道它并不是完全标准的，而且它可能会给移植带来问题。
另外，我认为正确答案之一可能是“对于较小的数据集更有效”，因为开销更小(是真的吗?)——因此，我希望将这个问题限制在键数量非普通的情况下(>1 024)。
编辑:哦，我忘记了显而易见的(感谢GMan!)——是的，地图当然是有序的——我知道，我正在寻找其他原因。
如果你想比较std::map和std::unordered_map实现的速度，你可以使用谷歌的sparsehash项目，它有一个time_hash_map程序来计时。例如，在x86_64 Linux系统上使用gcc 4.4.2
$ ./time_hash_map
TR1 UNORDERED_MAP (4 byte objects, 10000000 iterations):
map_grow 126.1 ns (27427396 hashes, 40000000 copies) 290.9 MB
map_predict/grow 67.4 ns (10000000 hashes, 40000000 copies) 232.8 MB
map_replace 22.3 ns (37427396 hashes, 40000000 copies)
map_fetch 16.3 ns (37427396 hashes, 40000000 copies)
map_fetch_empty 9.8 ns (10000000 hashes, 0 copies)
map_remove 49.1 ns (37427396 hashes, 40000000 copies)
map_toggle 86.1 ns (20000000 hashes, 40000000 copies)
STANDARD MAP (4 byte objects, 10000000 iterations):
map_grow 225.3 ns ( 0 hashes, 20000000 copies) 462.4 MB
map_predict/grow 225.1 ns ( 0 hashes, 20000000 copies) 462.6 MB
map_replace 151.2 ns ( 0 hashes, 20000000 copies)
map_fetch 156.0 ns ( 0 hashes, 20000000 copies)
map_fetch_empty 1.4 ns ( 0 hashes, 0 copies)
map_remove 141.0 ns ( 0 hashes, 20000000 copies)
map_toggle 67.3 ns ( 0 hashes, 20000000 copies)
