C#6.0的TryParse与出VAR参数 [英] C# 6.0 TryParse with out var param
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问题描述
在C#6.0新特性允许申报的TryParse方法内变量。
我有一些代码:
字符串s =你好;
如果(int.TryParse(S,出VAR结果))
{
}
我做错了什么?
PS:在项目设置C#6.0和.NET Framework 4.6的设置
解决方案
在C#6.0新特性允许申报的TryParse
方法中的变量。
块引用>
宣言表达式从C#6.0切在最终版并没有出货。目前,您不能这样做。 有它的C#7(见的:// github上.COM / DOTNET /罗斯林/问题/ 6183>此备查)。
A new feature in C# 6.0 allows to declare variable inside TryParse method. I have some code:
string s = "Hello"; if (int.TryParse(s, out var result)) { }
But I receive compile errors:
What I am doing wrong? P.S.: in project settings C# 6.0 and .NET framework 4.6 are set.
解决方案A new feature in C# 6.0 allows to declare variable inside TryParse method.
Declaration expressions was cut from C# 6.0 and wasn't shipped in the final release. You currently can't do that. There is a proposal for it on GitHub for C# 7 (also see this for future reference).
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