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Who knows Enumerable.Range(...)?

, 14 Apr 2013 CPOL
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Make loops behave deterministic-by-construction


I quite often see constructs like:

for(int n = 0; n < len; ++n)
   //... some more or less complex conditional control flow...
   n = 5; // causes the loop to never terminate if len is greater than 5
   //... some more code ...

This is a legal language construct, but it introduces the danger of non-determinism.

This tip aims to show some deterministic alternatives to that plain for loop.

Using the Code

If you need to loop over some range of integer, consider using the Enumerable.Range[^] based foreach loop instead:

using System.Linq;
foreach (int n in Enumerable.Range(0, len))

This guarantees that it iterates over all elements in strict sequence. Another benefit is that you cannot assign any other value to the loop variable n. E.g.

foreach (int n in Enumerable.Range(0, len))
    n = 5; // Compiler error!

Note: Enumerable.Range(...) is not from, to, but from, count:

// Summary:
//     Generates a sequence of integral numbers within a specified range.
// Parameters:
//   start:
//     The value of the first integer in the sequence.
//   count:
//     The number of sequential integers to generate.
// Returns:
//     An IEnumerable<Int32> in C# or IEnumerable(Of Int32) in Visual Basic that
//     contains a range of sequential integral numbers.
// Exceptions:
//   System.ArgumentOutOfRangeException:
//     count is less than 0.-or-start + count -1 is larger than System.Int32.MaxValue.
public static IEnumerable<int> Range(int start, int count);


One could also build his own iterator functions (see also C# Iterator Pattern demystified[^]). E.g.

public static IEnumerable<int> CountUp(int n, int count)
    while (count-- > 0) yield return n++;
public static IEnumerable<int> CountDown(int n, int count)
    while (count-- > 0) yield return n--;

When used like this...

foreach (int n in CountUp(0, 5)) Console.WriteLine(n);
foreach (int n in CountDown(100, 5)) Console.WriteLine(n);

...results in:


You can define any complexity of traversing sequence in that function and let the foreach-loop terminate deterministically, based on that sequence.


Try to avoid plain for (...) loops and replace by some deterministic loop alternative like:

  • foreach( ... Range(...))
  • foreach( ... CountUp(...))
  • etc.

Other alternatives are Linq iterations like Enumerable.Aggregate[^] etc. But these are a bit more advanced.


  • 2012-04-18 First version
  • 2012-04-19 Added hand-crafted CountUp/CountDown functions
  • 2013-04-14 Fixed some broken HTML markup in C# code generics


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


About the Author

Andreas Gieriet
Founder eXternSoft GmbH
Switzerland Switzerland
I feel comfortable on a variety of systems (UNIX, Windows, cross-compiled embedded systems, etc.) in a variety of languages, environments, and tools.
I have a particular affinity to computer language analysis, testing, as well as quality management.

More information about what I do for a living can be found at my LinkedIn Profile and on my company's web page (German only).

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Comments and Discussions

GeneralThoughts Pin
PIEBALDconsult14-Apr-13 15:30
memberPIEBALDconsult14-Apr-13 15:30 
GeneralMy vote of 5 Pin
Kenneth Haugland3-Sep-12 13:53
memberKenneth Haugland3-Sep-12 13:53 
GeneralRe: My vote of 5 Pin
Andreas Gieriet3-Sep-12 20:21
memberAndreas Gieriet3-Sep-12 20:21 
GeneralPros and Cons and ... Pin
Jani Giannoudis18-Apr-12 12:26
mvpJani Giannoudis18-Apr-12 12:26 
- deterministic
- Readability for non-numeric ranges

- Performance:[^]
- Readability for numeric ranges
- different pattern for ascending/descending iterations
- requires Linq

As a good coding practice, you never change the iteration variable within a loop.
Jani Giannoudis - Resource Sharing Made Easy | Co-founder

GeneralRe: Pros and Cons and ... Pin
Andreas Gieriet18-Apr-12 21:22
memberAndreas Gieriet18-Apr-12 21:22 
QuestionYour problem is not the for... Pin
Paulo Zemek18-Apr-12 9:48
mvpPaulo Zemek18-Apr-12 9:48 
AnswerRe: Your problem is not the for... Pin
Andreas Gieriet18-Apr-12 10:56
memberAndreas Gieriet18-Apr-12 10:56 
GeneralRe: Your problem is not the for... Pin
Paulo Zemek18-Apr-12 14:10
mvpPaulo Zemek18-Apr-12 14:10 
GeneralRe: Your problem is not the for... Pin
Andreas Gieriet18-Apr-12 18:05
memberAndreas Gieriet18-Apr-12 18:05 
AnswerRe: Your problem is not the for... Pin
TimothyP22-Apr-13 8:26
memberTimothyP22-Apr-13 8:26 

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