Visual Studio 2017的"实现接口"命令可以生成C#5代码吗？

Question

在Visual Studio 2017中,在" 快速操作和重构 "菜单中,当我单击" 实现接口 "时,我获得了使用C#的新功能的实现,如下所示:

public int ConnectionTimeout => throw new NotImplementedException();

问题是我的公司正在使用C#5,所以我们不能使用这种新的语法.我知道Visual Studio的旧版本会生成此代码:

public int ConnectionTimeout { get { throw new NotImplementedException(); } }

有没有办法让Visual Studio 2017执行此操作？我已经将我的项目设置为使用C#5进行编译.

编辑:我注意到代码生成器不一致.如果接口具有仅具有getter的bool成员,则它使用旧语法.否则它使用新语法.这让我觉得我运气不好.

private interface myint
{
    bool bool1 { get; }
    bool bool2 { get; set; }
    bool bool3 { set; }

    int int1 { get; }
    int int2 { get; set; }
    int int3 { set; }

    string string1 { get; }
    string string2 { get; set; }
    string string3 { set; }
}

private class myclass : myint //I clicked "Implement Interface" on this
{
    public bool bool1
    {
        get
        {
            throw new NotImplementedException();
        }
    }

    public bool bool2 { get => throw new NotImplementedException(); set => throw new NotImplementedException(); }
    public bool bool3 { set => throw new NotImplementedException(); }

    public int int1 => throw new NotImplementedException();

    public int int2 { get => throw new NotImplementedException(); set => throw new NotImplementedException(); }
    public int int3 { set => throw new NotImplementedException(); }

    public string string1 => throw new NotImplementedException();

    public string string2 { get => throw new NotImplementedException(); set => throw new NotImplementedException(); }
    public string string3 { set => throw new NotImplementedException(); }
}

Answer 1

转到工具 - >选项 - >文本编辑器 - >代码样式 - >常规,滚动到代码块首选项部分并更改首选项对于属性,对于索引器和对于访问者从"首选表达式块"(默认)更改为"首选块体" ".