如何处理返回Task<Something>的异步方法中的异常

try
{
    Something result = await SomeAsyncMethod();
    // 处理返回结果
}
catch (Exception ex)
{
    // 处理异常
}

Task<Something> task = SomeAsyncMethod();
await task;

if (task.Exception != null)
{
    // 处理异常
}
else
{
    Something result = task.Result;
    // 处理返回结果
}

Task<Something> task = SomeAsyncMethod();
task.ContinueWith(t =>
{
    if (t.Exception != null)
    {
        // 处理异常
    }
    else
    {
        Something result = t.Result;
        // 处理返回结果
    }
});