在实现接口时删除异步/等待警告CS1998

发布于 2025-02-12 18:32:00 字数 856 浏览 0 评论 0原文

假设我实现了有两种方法的接口：创建和createAsync。在我的情况下，异步方法不是真正异步，它使用Syncronous 创建方法返回创建对象：

public Task<GrandeurDTO> CreateAsync()
{
    var dto = Create();
    return Task.FromResult(dto);
}
public GrandeurDTO Create()
{
    var bo = new Grandeur();

    var dto = _mapper.Map<GrandeurDTO>(bo);
    return dto;
}

我应该如何删除警告

这种异步方法缺乏“等待”运算符，并且会同步运行。考虑使用“等待”操作员等待非阻止API调用，或'等待任务。lun（CS1998）

知道我不能在create中重命名该方法（我已经有一个;原因；如果不是一个问题，则与这个）

原文

Suppose I implement an interface where there are two methods: Create and CreateAsync. The Async method in my case is not really async, and it uses the syncronous Create method to return the created object:

public Task<GrandeurDTO> CreateAsync()
{
    var dto = Create();
    return Task.FromResult(dto);
}
public GrandeurDTO Create()
{
    var bo = new Grandeur();

    var dto = _mapper.Map<GrandeurDTO>(bo);
    return dto;
}

How should I remove the warning

This async method lacks 'await' operators and will run synchronously.
Consider using the 'await' operator to await non-blocking API calls,
or 'await Task.Run(CS1998)

knowing I can't rename the method in Create (I already have one; cause if not the question is similar to this one)