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233

answers:

3

Hi i have a program that deals alot with vectors and indexes of the elements of these vectors , and I was wondering :

  1. is there a difference between uint and unsigned int
  2. which is better to use one of the above types or just use "int" as I read some people say compiler does handle int values more efficiently , but if i used int i will have to check always for negative idxs which is pain.
  3. do you think iterators to be better ? is it more efficient than normal indexing vectorx[idx] ?

p.s the software will handle large data processes and good performance is a must have requirement Thanks alot

+4  A: 
  1. C++ defines no such type as uint. This must be "your" type, i.e. a type defined in your code or some third party library. One can guess that it is the same as unsigned int. Could be unsigned long int though or something else. Anyway, you have to check it yourself.

  2. It is a matter of personal style. I, for example, believe that one has to use unsigned types to represent naturally non-negative values, like sizes or quantities. There's no difference in performance between signed and unsigned types, aside from some specific contexts. I would say that in most cases it is unsigned types that will be handled more efficiently.

  3. Iterators make implementations more generic, i.e. you can use sequential-access iterator and thus make your implementation applicable to any sequential data structure. By using index you impose the random-access requirement on the data structure, which is a strong requirement. It is not a good idea to impose strong requirements when there's no real need for them.

AndreyT
ismail marmoush
`uint` is typedef'd as a shorthand to `unsigned int` by most compilers.
You
In any case, it doesn't change the fact that theres no such type as `uint` neither in C nor in C++, so unless a specific compiler is considered, question about `uint` have no meaningful answer.
AndreyT
@AndreyT, I'd also add you shouldn't use either `unsigned int` or `unsigned long` when dealing with sizes of containers. At least in C++, as the question is tagged, you should be using `size_t`.
Nathan Ernst
@Nathan Ernst: Not true. You are talking about containers in general, right? You should *never* use `size_t` with containers in general. There's no guarantee that the range of `size_t` is sufficient for any container. `size_t` is only sufficient with array-based containers, but noting else. Also, when dealing with standard containers, you should use `container_type::size_type` and not `size_t`.
AndreyT
Finally, the appropriate type for each given *specific* context is dictated by that specific context, i.e. your specific application. Could easily be `unsigned int` or `unsigned long`. `size_t` is only appropriate in generic array-based contexts.
AndreyT
+2  A: 

If you're looping through the vector sequentially, by all means, use the iterator. There is overhead related to indexing, regardless of the index type, which can be avoided by iterating.

TreDubZedd
A: 

1) uint = unsigned int, in fact uint is just a typedef for unsigned int (will be replaced by unsigned int on compile time).

2) If you want to add to your code some "security" go with uint, you'll avoid for sure negative values.

3) If you run through the vector sequentially, go with iterators, they are optimized for sequential looping (they are some kind of pointers).

Ars
Not bad enough for a downvote, but `uint` will usually be a `typedef`, not a macro, and signed versus unsigned has little to do with security (if anything it may cause security bugs, since unsigned arithmetic works differently than many programmers expect!)
Tyler McHenry
You're right for typedef (I'll modify it)By security, I mean a security feeling for the programmer, nothing related to the compiler, that's why I've put the ""
Ars