c# - Create deferred enumerable from Func<IEnumerable<T>> -

is there built-in/standard way of creating deferred ienumerable<t>, given func<ienumerable<t>>? google skills may weak...

suppose know method foo returns ienumerable<t> executes (and perhaps long-running). want pass method bar takes in ienumerable<t> don't want foo execute long-running process return items until bar starts iterating on items:

ienumerable<string> foo() {     var items = getitems(); // long running process     return items; }  void bar(ienumerable<string> foo) {      /* else long-running, or return if pre-condition fails , don't want iterate on foo */     foreach (var item in foo)     {         //     } }     void run() {     ienumerable<string> foo = getfoowrapper(foo)     bar(items); }  ienumerable<string> getfoowrapper(func<ienumerable<string>> foo) {     // ??? } 

what's best way of implementing getfoowrapper? simplest approach following, if there's "better" way of doing it, i'd know it. not matters, r# suggests simplifying return foo();.

ienumerable<string> getfoowrapper(func<ienumerable<string>> foo) {     foreach (var item in foo())     {         yield return item;     } } 

sounds you're looking lazy<t> class. defers creation of object until first used.

here's example defer execution of foo method until value property accessed first time. once foo executed once, not executed again unless new lazy<t> created.

var items = new lazy<ienumerable<string>>(foo); foreach(var item in items.value) // delay starts here {     // work }