[英]Resharper C++ enforce C99
I am writing an embedded application in plain C99
in Visual Studio with VisualGDB and Resharper c++. 我正在使用VisualGDB和Resharper c ++在Visual Studio的纯
C99
中编写嵌入式应用程序。 The Resharper website states that it fully supports C99
. Resharper网站指出,它完全支持
C99
。
But if I have a typedef struct like this 但是如果我有一个像这样的typedef结构
typedef struct { int Bar;} Foo_t;
And some function that return this struct 还有一些返回此结构的函数
Foo_t Foo(void){
return (Foo_t){.Bar = 0;};
}
Resharper is completely confused (unreachable code, missing semicolon etc.). Resharper完全感到困惑(无法访问的代码,缺少分号等)。 Is there any way to turn on
C99
standard in resharper or does it just not support this type of returns? 是否可以通过任何方式打开
C99
标准,还是不支持这种类型的退货?
We are sorry for the inconvenience. 我们对造成的不便很抱歉。 I've created an issue in YouTrack about it: https://youtrack.jetbrains.com/issue/RSCPP-15938 You can follow this request and vote for it.
我已经在YouTrack中创建了一个与此有关的问题: https ://youtrack.jetbrains.com/issue/RSCPP-15938您可以按照此请求进行投票。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.