Is it defined to provide an empty range to C++ standard algorithms?

问题

Following on from my previous question, can we prove that the standard allows us to pass an empty range to a standard algorithm?

Paragraph 24.1/7 defines an "empty range" as the range [i,i) (where i is valid), and i would appear to be "reachable" from itself, but I'm not sure that this qualifies as a proof.

In particular, we run into trouble when looking at the sorting functions. For example, std::sort:

Complexity: O(N log(N)) (where N == last - first) comparisons

Since log(0) is generally considered to be undefined, and I don't know what 0*undefined is, could there be a problem here?

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^{(Yes, ok, I'm being a bit pedantic. Of course no self-respecting stdlib implementation would cause a practical problem with an empty range passed to std::sort. But I'm wondering whether there's a potential hole in the standard wording here.)}

回答1:

I don't seem much room for question. In §24.1/6 we're told:

An iterator j is called reachable from an iterator i if and only if there is a finite sequence of applications of the expression ++i that makes i == j.

and in $24.1/7:

Range [i, j) is valid if and only if j is reachable from i.

Since 0 is finite, [i, i) is a valid range. §24.1/7 goes on to say:

The result of the application of functions in the library to invalid ranges is undefined.

That doesn't go quite so far as to say that a valid range guarantees defined results (reasonable, since there are other requirements, such as on the comparison function) but certainly seems to imply that a range being empty, in itself, should not lead to UB or anything like that. In particular, however, the standard makes an empty range just another valid range; there's no real differentiation between empty and non-empty valid ranges, so what applies to a non-empty valid range applies equally well to an empty valid range.

回答2:

Big-O notation is defined in terms of the limit of the function. An algorithm with actual running time g(N) is O(f(N)) if and only if lim N→∞ g(N)/f(N) is a non-negative real number g(N)/f(N) is less than some positive real number C for all values N greater than some constant k (the exact values of C and k are immaterial; you just have to be able to find any C and k that makes this true). (thanks for the correction, Jesse!)

You'll note that the actual number of elements is not relevant in big-O analysis. Big-O analysis says nothing about the behavior of the algorithm for small numbers of elements; therefore, it does not matter if f(N) is defined at N=0. More importantly, the actual runtime behavior is controlled by a different function g(N), which may well be defined at N=0 even if f(0) is undefined.

回答3:

Apart from the relevant answer given by @bdonlan, note also that f(n) = n * log(n) does have a well-defined limit as n goes to zero, namely 0. This is because the logarithm diverges more slowly than any polynomial, in particular, slower than n. So all is well :-)

来源：https://stackoverflow.com/questions/7505267/is-it-defined-to-provide-an-empty-range-to-c-standard-algorithms