Why can't "return" and "yield return" be used in the same method?

Question

Why can't we use both return and yield return in the same method?

For example, we can have GetIntegers1 and GetIntegers2 below, but not GetIntegers3.

public IEnumerable<int> GetIntegers1() {   return new[] { 4, 5, 6 }; }  public IEnumerable<int> GetIntegers2() {   yield return 1;   yield return 2;   yield return 3; }  public IEnumerable<int> GetIntegers3() {   if ( someCondition )   {     return new[] {4, 5, 6}; // compiler error   }   else   {     yield return 1;     yield return 2;     yield return 3;   } } 

Answer 1

return is eager. It returns the entire resultset at once. yield return builds an enumerator. Behind the scenes the C# compiler emits the necessary class for the enumerator when you use yield return. The compiler doesn't look for runtime conditions such as if ( someCondition ) when determining whether it should emit the code for an enumerable or have a method that returns a simple array. It detects that in your method you are using both which is not possible as he cannot emit the code for an enumerator and at the same time have the method return a normal array and all this for the same method.

Answer 2

No you can't do that - an iterator block (something with a yield) cannot use the regular (non-yield) return. Instead, you need to use 2 methods:

public IEnumerable<int> GetIntegers3() {   if ( someCondition )   {     return new[] {4, 5, 6}; // compiler error   }   else   {     return GetIntegers3Deferred();   } } private IEnumerable<int> GetIntegers3Deferred() {     yield return 1;     yield return 2;     yield return 3; } 

or since in this specific case the code for both already exists in the other 2 methods:

public IEnumerable<int> GetIntegers3() {   return ( someCondition ) ? GetIntegers1() : GetIntegers2(); }