c# - 检查 OpenFileDialog.OpenFile 是否为空

Question

我已经在 the msdn doc 中查看了这段代码:

Stream myStream = null;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
[...] // Some init
try
{
    if ((myStream = openFileDialog1.OpenFile()) != null)
    {
        using (myStream)
        {
            // Insert code to read the stream here.
        }
    }
}

但是 Resharper 温柔地告诉我检查 null 是没有用的:

我应该相信 Resharper 还是微软？

Answer 1

这是(大部分)OpenFile() 的源代码.

try block 有可能抛出异常，并且该方法返回 null 的 Stream，所以我会相信 MSDN 文档。我不确定 ReSharper 为何提出该建议。

public Stream OpenFile()
{ 
    string filename = FileNamesInternal[0];

    if (filename == null || (filename.Length == 0))
        throw new ArgumentNullException("FileName");

    Stream s = null; 

    new FileIOPermission(FileIOPermissionAccess.Read, IntSecurity.UnsafeGetFullPath(filename)).Assert();
    try
    {
        s = new FileStream(filename, FileMode.Open, FileAccess.Read, FileShare.Read); 
    }
    finally 
    { 
        CodeAccessPermission.RevertAssert();
    } 
    return s;
}