why does the compiler NOT complain about not all paths returning?
I have this inside of a do..while:
yield return string.Join(",", arr) + "n";
Why isn't the compiler complaining that not all of the code paths are returning a value?
The full code example is below:
public static IEnumerable<string> Convert(Stream stream)
{
System.Text.Encoding.RegisterProvider(System.Text.CodePagesEncodingProvider.Instance);
IExcelDataReader reader = ExcelReaderFactory.CreateBinaryReader(stream);
var csvContent = string.Empty;
do
{
while (reader.Read())
{
var arr = new List<string>();
for (int i = 0; i < reader.FieldCount; i++)
{
var cell = reader[i]?.ToString();
var format = reader.GetNumberFormatString(i);
if (format == "mm\/dd\/yyyy" || format == "M/d/yyyy")
{
cell = cell.Replace(" 12:00:00 AM", "");
}
if (format == "h\:mm\:ss AM/PM")
{
cell = cell.Replace("12/31/1899 ", "");
}
var processedCell = cell == null ? string.Empty : cell.Contains(",") ? """ + cell + """ : cell;
arr.Add(processedCell);
}
yield return string.Join(",", arr) + "n";
}
} while (reader.NextResult());
}
There is no return keyword as the last line!
c# .net visual-studio visual-studio-2017
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
|
show 2 more comments
I have this inside of a do..while:
yield return string.Join(",", arr) + "n";
Why isn't the compiler complaining that not all of the code paths are returning a value?
The full code example is below:
public static IEnumerable<string> Convert(Stream stream)
{
System.Text.Encoding.RegisterProvider(System.Text.CodePagesEncodingProvider.Instance);
IExcelDataReader reader = ExcelReaderFactory.CreateBinaryReader(stream);
var csvContent = string.Empty;
do
{
while (reader.Read())
{
var arr = new List<string>();
for (int i = 0; i < reader.FieldCount; i++)
{
var cell = reader[i]?.ToString();
var format = reader.GetNumberFormatString(i);
if (format == "mm\/dd\/yyyy" || format == "M/d/yyyy")
{
cell = cell.Replace(" 12:00:00 AM", "");
}
if (format == "h\:mm\:ss AM/PM")
{
cell = cell.Replace("12/31/1899 ", "");
}
var processedCell = cell == null ? string.Empty : cell.Contains(",") ? """ + cell + """ : cell;
arr.Add(processedCell);
}
yield return string.Join(",", arr) + "n";
}
} while (reader.NextResult());
}
There is no return keyword as the last line!
c# .net visual-studio visual-studio-2017
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
@DavidG exactly! and my signature is promising anIEnumerable<string>return value, not avoid!
– l--''''''---------''''''''''''
Nov 23 at 1:27
1
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
3
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
2
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36
|
show 2 more comments
I have this inside of a do..while:
yield return string.Join(",", arr) + "n";
Why isn't the compiler complaining that not all of the code paths are returning a value?
The full code example is below:
public static IEnumerable<string> Convert(Stream stream)
{
System.Text.Encoding.RegisterProvider(System.Text.CodePagesEncodingProvider.Instance);
IExcelDataReader reader = ExcelReaderFactory.CreateBinaryReader(stream);
var csvContent = string.Empty;
do
{
while (reader.Read())
{
var arr = new List<string>();
for (int i = 0; i < reader.FieldCount; i++)
{
var cell = reader[i]?.ToString();
var format = reader.GetNumberFormatString(i);
if (format == "mm\/dd\/yyyy" || format == "M/d/yyyy")
{
cell = cell.Replace(" 12:00:00 AM", "");
}
if (format == "h\:mm\:ss AM/PM")
{
cell = cell.Replace("12/31/1899 ", "");
}
var processedCell = cell == null ? string.Empty : cell.Contains(",") ? """ + cell + """ : cell;
arr.Add(processedCell);
}
yield return string.Join(",", arr) + "n";
}
} while (reader.NextResult());
}
There is no return keyword as the last line!
c# .net visual-studio visual-studio-2017
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
I have this inside of a do..while:
yield return string.Join(",", arr) + "n";
Why isn't the compiler complaining that not all of the code paths are returning a value?
The full code example is below:
public static IEnumerable<string> Convert(Stream stream)
{
System.Text.Encoding.RegisterProvider(System.Text.CodePagesEncodingProvider.Instance);
IExcelDataReader reader = ExcelReaderFactory.CreateBinaryReader(stream);
var csvContent = string.Empty;
do
{
while (reader.Read())
{
var arr = new List<string>();
for (int i = 0; i < reader.FieldCount; i++)
{
var cell = reader[i]?.ToString();
var format = reader.GetNumberFormatString(i);
if (format == "mm\/dd\/yyyy" || format == "M/d/yyyy")
{
cell = cell.Replace(" 12:00:00 AM", "");
}
if (format == "h\:mm\:ss AM/PM")
{
cell = cell.Replace("12/31/1899 ", "");
}
var processedCell = cell == null ? string.Empty : cell.Contains(",") ? """ + cell + """ : cell;
arr.Add(processedCell);
}
yield return string.Join(",", arr) + "n";
}
} while (reader.NextResult());
}
There is no return keyword as the last line!
c# .net visual-studio visual-studio-2017
c# .net visual-studio visual-studio-2017
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
edited Nov 23 at 1:28
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
asked Nov 23 at 1:23
l--''''''---------''''''''''''
9,868232532876
9,868232532876
@DavidG exactly! and my signature is promising anIEnumerable<string>return value, not avoid!
– l--''''''---------''''''''''''
Nov 23 at 1:27
1
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
3
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
2
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36
|
show 2 more comments
@DavidG exactly! and my signature is promising anIEnumerable<string>return value, not avoid!
– l--''''''---------''''''''''''
Nov 23 at 1:27
1
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
3
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
2
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36
@DavidG exactly! and my signature is promising an
IEnumerable<string> return value, not a void !– l--''''''---------''''''''''''
Nov 23 at 1:27
@DavidG exactly! and my signature is promising an
IEnumerable<string> return value, not a void !– l--''''''---------''''''''''''
Nov 23 at 1:27
1
1
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
3
3
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
2
2
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36
|
show 2 more comments
2 Answers
2
active
oldest
votes
Because you are using yield return. yield return method ends not with return, but actually with yield break and last yield break can be omitted
UPD. This is example of old code (Robotics Studio and Unity) where yield break on last line is exists
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
add a comment |
This is an exception in connection with the IEnumerable iterator pattern and the yield return keyword. In this case, the compiler constructs its state machine for the iterator runtime, and does not apply the same set of path coverage as it does with linear methods.
The semantics goes, "an iterator that does not produce, is empty, and this is at the same time a valid substitution for any code path that does not explicitely return a value".
The key to understanding this is that the compiler reforms methods that constitute iterators, to code that does no longer have uninitialized return values. See C# standard chapter 10.4.4.1 for an example.
Related: https://stackoverflow.com/a/9631242/1132334
answered Nov 23 at 1:34
dlatikay
6,22521848
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
add a comment |
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2 Answers
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2 Answers
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Because you are using yield return. yield return method ends not with return, but actually with yield break and last yield break can be omitted
UPD. This is example of old code (Robotics Studio and Unity) where yield break on last line is exists
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
add a comment |
Because you are using yield return. yield return method ends not with return, but actually with yield break and last yield break can be omitted
UPD. This is example of old code (Robotics Studio and Unity) where yield break on last line is exists
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
add a comment |
Because you are using yield return. yield return method ends not with return, but actually with yield break and last yield break can be omitted
UPD. This is example of old code (Robotics Studio and Unity) where yield break on last line is exists
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
Because you are using yield return. yield return method ends not with return, but actually with yield break and last yield break can be omitted
UPD. This is example of old code (Robotics Studio and Unity) where yield break on last line is exists
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
edited Nov 23 at 1:59
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
answered Nov 23 at 1:40
Alexei Shcherbakov
28225
28225
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
add a comment |
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
1
1
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
"last yield break can be omitted" - What? No. That's not right.
– Enigmativity
Nov 23 at 1:41
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
I've updated answer with code sample from book ~2008 year.
– Alexei Shcherbakov
Nov 23 at 1:56
2
2
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
Thanks for posting an answer. Please beware that pictures of code are frowned upon on SO because one can't copy/paste to try it. Could you please update your answer?
– Jacques Gaudin
Nov 23 at 8:30
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
How do you qualify “frowned upon” when this is the accepted answer?
– l--''''''---------''''''''''''
Nov 23 at 11:04
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
first code sample is from copyrighted book and second was founded by google image search. I remember that 'yield break' was necessary for Visual Studio more than 10 years ago (I even can't remember - Visual Studio 2005 Beta 2 or RTM), so this is only "historical" sample - in modern version of C# language it is not necessary. So I tried to find examples of 'yield break' in Internet in pre2005-2008 years :-)
– Alexei Shcherbakov
Nov 23 at 20:18
add a comment |
This is an exception in connection with the IEnumerable iterator pattern and the yield return keyword. In this case, the compiler constructs its state machine for the iterator runtime, and does not apply the same set of path coverage as it does with linear methods.
The semantics goes, "an iterator that does not produce, is empty, and this is at the same time a valid substitution for any code path that does not explicitely return a value".
The key to understanding this is that the compiler reforms methods that constitute iterators, to code that does no longer have uninitialized return values. See C# standard chapter 10.4.4.1 for an example.
Related: https://stackoverflow.com/a/9631242/1132334
answered Nov 23 at 1:34
dlatikay
6,22521848
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
add a comment |
This is an exception in connection with the IEnumerable iterator pattern and the yield return keyword. In this case, the compiler constructs its state machine for the iterator runtime, and does not apply the same set of path coverage as it does with linear methods.
The semantics goes, "an iterator that does not produce, is empty, and this is at the same time a valid substitution for any code path that does not explicitely return a value".
The key to understanding this is that the compiler reforms methods that constitute iterators, to code that does no longer have uninitialized return values. See C# standard chapter 10.4.4.1 for an example.
Related: https://stackoverflow.com/a/9631242/1132334
answered Nov 23 at 1:34
dlatikay
6,22521848
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
add a comment |
This is an exception in connection with the IEnumerable iterator pattern and the yield return keyword. In this case, the compiler constructs its state machine for the iterator runtime, and does not apply the same set of path coverage as it does with linear methods.
The semantics goes, "an iterator that does not produce, is empty, and this is at the same time a valid substitution for any code path that does not explicitely return a value".
The key to understanding this is that the compiler reforms methods that constitute iterators, to code that does no longer have uninitialized return values. See C# standard chapter 10.4.4.1 for an example.
Related: https://stackoverflow.com/a/9631242/1132334
answered Nov 23 at 1:34
dlatikay
6,22521848
This is an exception in connection with the IEnumerable iterator pattern and the yield return keyword. In this case, the compiler constructs its state machine for the iterator runtime, and does not apply the same set of path coverage as it does with linear methods.
The semantics goes, "an iterator that does not produce, is empty, and this is at the same time a valid substitution for any code path that does not explicitely return a value".
The key to understanding this is that the compiler reforms methods that constitute iterators, to code that does no longer have uninitialized return values. See C# standard chapter 10.4.4.1 for an example.
Related: https://stackoverflow.com/a/9631242/1132334
answered Nov 23 at 1:34
dlatikay
6,22521848
edited Nov 23 at 1:44
answered Nov 23 at 1:34
dlatikay
6,22521848
answered Nov 23 at 1:34
dlatikay
6,22521848
answered Nov 23 at 1:34
dlatikay
6,22521848
6,22521848
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
add a comment |
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
another words, you are in agreement with hans?
– l--''''''---------''''''''''''
Nov 23 at 1:36
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
we both are in agreement with the specification. only that Hans obviously knows it by heart, and I have to look it up.
– dlatikay
Nov 23 at 1:45
add a comment |
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@DavidG exactly! and my signature is promising an
IEnumerable<string>return value, not avoid!– l--''''''---------''''''''''''
Nov 23 at 1:27
1
That is the nature of yield return. Check example from Microsoft. docs.microsoft.com/en-us/dotnet/csharp/language-reference/… If exponent is 0, it is the same as yours.
– wannadream
Nov 23 at 1:29
3
Iterators march to a different drummer. Not returning anything is just fine, it returns an empty iterator.
– Hans Passant
Nov 23 at 1:31
@HansPassant very nice explanation, but where does it say that?
– l--''''''---------''''''''''''
Nov 23 at 1:32
2
The C# language specification, chapter 10.4.4.1. It doesn't exactly yell it out.
– Hans Passant
Nov 23 at 1:36