Can one iterate on [0, n-1] interval with C++11 range-based-for?

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Is there a range class in C++11 for use with range based for loops?

I.e. is there a standard range with iterator that will dereference to an integer? I'm thinking about something like this:

for (int i : rangeTo(10)) { ... }
for (int i : rangeFromTo(10, 20)) { .... }

What is the advantage of that over `for(int i = 10; i < 20; ++i)`? — Daniel Fischer, Jul 23 '12 at 23:54
Łukasz, no difference. Well, actually there is. With the classical for loop, I know whether I need a strict or weak inequality. With a range, I'd always have to check whether that includes the upper bound or not, as in Python. — Daniel Fischer, Jul 24 '12 at 01:22
@DanielFischer: Nonsense. I have yet to see any C or C++ system or library (which *isn't* interfacing with some scripting language) that doesn't use a half-open range. It is absolutely standard in C/C++. Standard algorithms do it. Boost.Range does it. Virtually every for loop over some range of numbers does it. It's so common that if you're not using a half-open range in C/C++, it's usually considered a code smell. — Nicol Bolas, Jul 24 '12 at 01:36
@NicolBolas Ah, then I didn't get the inequality wrong? I thought I had, per Łukasz's comment. And that would confuse me, since half open is what I'd expect. — Daniel Fischer, Jul 24 '12 at 01:41
@Daniel - You did get it slightly wrong. :-) Łukasz wants you to use `!=` for inequality, not `<`. We are picky around here! — Bo Persson, Jul 24 '12 at 12:28

Jesse Good · Accepted Answer · 2012-07-24T00:01:17.720

6

No, but there is boost::irange:

#include <boost/range/irange.hpp>
...
for (auto i : boost::irange(10, 20))
{
    std::cout << i << ' ';
}

and boost::counting_range

for (auto i : boost::counting_range(10, 20))
{
    std::cout << i << ' ';
}

The difference is you can add a step, for boost::irange(10, 20, 2).

edited Jul 24 '12 at 00:01

answered Jul 23 '12 at 23:56

Jesse Good

1 Answers1