FileStream:The process cannot access the file because it is being used by another process

先看下面一段代码（先以共享的方式打开文件读写，然后以只读的方式打开相同文件）：

FileStream fs = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite);
FileStream fs2 = new FileStream(filePath, FileMode.Open, FileAccess.Read) 或者 new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Read);

第一句成功执行，第二句呢？它抛出访问违规异常:The process cannot access the file 'c:\Odma32.log' because it is being used by another process！

查阅MSDN，无果。后经多次实验，包括使用Windows API CreateFile(...),最终发现正确的用法居然是：

FileStream fs = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite);
FileStream fs2 = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.ReadWrite);

在打开fs2时,它必须制定FileShare为ReadWrite，因为fs以ReadWrite的方式打开的。

结论：FileShare不只是对随后的打开文件请求有影响，但事实是它对已经打开的文件句柄也有影响。MSDN中关于FileShare的解释不到位，应该CreateFile条目中的这一段也放进去：

You cannot request a sharing mode that conflicts with the access mode that is specified in an existing request that has an open handle. CreateFile would fail and the GetLastError function would return ERROR_SHARING_VIOLATION.

转载于:https://www.cnblogs.com/mschen/p/5353862.html

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