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c++ - Jump Table Switch Case question

I am trying to understand some things about jump tables and its relationship between a switch case statement.

I was told that a jump table is a O(1) structure that the compiler generates which makes lookup of values essentially about as fast as you can get. However in some cases a Hashtable/Dictionary might be faster. I was also told this will only work if the switch case contains ordered data values.

Can someone please confirm or deny this and explain what a jump table is, it's importance and the time complexity versus using a dictionary or hashtable. Thanks.

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A jump table is an abstract structure used to transfer control to another location. Goto, continue, and break are similar, except they always transfer to a specific location instead of one possibility from many. In particular, this control flow is not the same as a function call. (Wikipedia's article on branch tables is related.)

A switch statement is how to write jump tables in C/C++. Only a limited form is provided (can only switch on integral types) to make implementations easier and faster in this common case. (How to implement jump tables efficiently has been studied much more for integral types than for the general case.) A classic example is Duff's Device.

However, the full capability of a jump table is often not required, such as when every case would have a break statement. These "limited jump tables" are a different pattern, which is only taking advantage of a jump table's well-studied efficiency, and are common when each "action" is independent of the others.


Actual implementations of jump tables take different forms, mostly differing in how the key to index mapping is done. That mapping is where terms like "dictionary" and "hash table" come in, and those techniques can be used independently of a jump table. Saying that some code "uses a jump table" doesn't imply by itself that you have O(1) lookup.

The compiler is free to choose the lookup method for each switch statement, and there is no guarantee you'll get one particular implementation; however, compiler options such as optimize-for-speed and optimize-for-size should be taken into account.

You should look into studying data structures to get a handle on the different complexity requirements imposed by them. Briefly, if by "dictionary" you mean a balanced binary tree, then it is O(log n); and a hash table depends on its hash function and collision strategy. In the particular case of switch statements, since the compiler has full information, it can generate a perfect hash function which means O(1) lookup. However, don't get lost by just looking at overall algorithmic complexity: it hides important factors.


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