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c# - File.Create 和 File.OpenRead 和 IOException

转载 作者:行者123 更新时间:2023-11-30 14:38:45 26 4
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我需要读取一个在程序的其他部分打开用于写入的文件

const string fileName = "file.bin";
FileStream create = File.Open(fileName, FileMode.Create, FileAccess.Write, FileShare.Read);
FileStream openRead = File.Open(fileName, FileMode.Open, FileAccess.Read, FileShare.Read);

最后一行引发 IOException:

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

请帮助正确配置 File.Open 参数。

最佳答案

在两个语句中将 FileShare 参数更改为 FileShare.ReadWrite:

FileStream create = File.Open(fileName, FileMode.Create, FileAccess.Write, FileShare.ReadWrite);
FileStream openRead = File.Open(fileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite);

ReadWrite 标志描述来自 MSDN:

Allows subsequent opening of the file for reading or writing. If this flag is not specified, any request to open the file for reading or writing (by this process or another process) will fail until the file is closed.

关于c# - File.Create 和 File.OpenRead 和 IOException,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7428417/

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