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Alternative to out variables
What's the difference between the 'ref' and 'out' keywords?Parse JSON in C#Performance surprise with “as” and nullable typesIs there a reason for C#'s reuse of the variable in a foreach?Does anyone have benchmarks (code & results) comparing performance of Android apps written in Xamarin C# and Java?Why not inherit from List<T>?How to upgrade a project to C# 6Why can I assign 0.0 to enumeration values, but not 1.0How to use C# 6 with Web Site project type?TryParse with out var param
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.
c#
asked Mar 8 at 7:07
HoseHose
626
add a comment |
I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.
c#
asked Mar 8 at 7:07
HoseHose
626
5
Just declare the variable before using it inTryParse.double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);
– Paweł Łukasik
Mar 8 at 7:08
1
Got it, thanks.
– Hose
Mar 8 at 8:10
add a comment |
I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.
c#
asked Mar 8 at 7:07
HoseHose
626
I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.
c#
c#
asked Mar 8 at 7:07
HoseHose
626
asked Mar 8 at 7:07
HoseHose
626
asked Mar 8 at 7:07
HoseHose
626
asked Mar 8 at 7:07
HoseHose
626
asked Mar 8 at 7:07
HoseHose
626
626
5
Just declare the variable before using it inTryParse.double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);
– Paweł Łukasik
Mar 8 at 7:08
1
Got it, thanks.
– Hose
Mar 8 at 8:10
add a comment |
5
Just declare the variable before using it inTryParse.double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);
– Paweł Łukasik
Mar 8 at 7:08
1
Got it, thanks.
– Hose
Mar 8 at 8:10
5
5
Just declare the variable before using it in
TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);– Paweł Łukasik
Mar 8 at 7:08
Just declare the variable before using it in
TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);– Paweł Łukasik
Mar 8 at 7:08
1
1
Got it, thanks.
– Hose
Mar 8 at 8:10
Got it, thanks.
– Hose
Mar 8 at 8:10
add a comment |
2 Answers
2
active
oldest
votes
You don't need to inline declare a type for out-parameters.
Replace:
double.TryParse(wordConf, out double wordConfDouble);
With:
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
add a comment |
It's just the inline declaration which is not supported in < C#7.0. Change your code to
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
Reference: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables
answered Mar 8 at 7:09
fubofubo
31k970107
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You don't need to inline declare a type for out-parameters.
Replace:
double.TryParse(wordConf, out double wordConfDouble);
With:
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
add a comment |
You don't need to inline declare a type for out-parameters.
Replace:
double.TryParse(wordConf, out double wordConfDouble);
With:
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
add a comment |
You don't need to inline declare a type for out-parameters.
Replace:
double.TryParse(wordConf, out double wordConfDouble);
With:
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
You don't need to inline declare a type for out-parameters.
Replace:
double.TryParse(wordConf, out double wordConfDouble);
With:
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
answered Mar 8 at 7:09
Eric McLachlanEric McLachlan
10617
10617
add a comment |
add a comment |
It's just the inline declaration which is not supported in < C#7.0. Change your code to
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
Reference: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables
answered Mar 8 at 7:09
fubofubo
31k970107
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
add a comment |
It's just the inline declaration which is not supported in < C#7.0. Change your code to
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
Reference: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables
answered Mar 8 at 7:09
fubofubo
31k970107
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
add a comment |
It's just the inline declaration which is not supported in < C#7.0. Change your code to
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
Reference: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables
answered Mar 8 at 7:09
fubofubo
31k970107
It's just the inline declaration which is not supported in < C#7.0. Change your code to
double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);
Reference: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables
answered Mar 8 at 7:09
fubofubo
31k970107
answered Mar 8 at 7:09
fubofubo
31k970107
answered Mar 8 at 7:09
fubofubo
31k970107
answered Mar 8 at 7:09
fubofubo
31k970107
31k970107
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
add a comment |
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
2
2
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.
– Hose
Mar 8 at 8:10
add a comment |
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5
Just declare the variable before using it in
TryParse.double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);– Paweł Łukasik
Mar 8 at 7:08
1
Got it, thanks.
– Hose
Mar 8 at 8:10