File.Open(String, FileMode, FileAccess, FileShare) Method in C# with Examples
File.Open(String, FileMode, FileAccess, FileShare) is an inbuilt File class method that is used to open a FileStream on the specified path, having the specified mode with read, write, or read/write access and the specified sharing option.
Syntax:
public static System.IO.FileStream Open (string path, System.IO.FileMode mode, System.IO.FileAccess access, System.IO.FileShare share);
Parameter: This function accepts three parameters which are illustrated below:
- path: This is the specified file to open.
- mode: This mode value specifies whether a new file is created if one does not exist, and also determines whether the existing file’s contents are retained or overwritten.
- access: This value specifies the operations that can be performed on the file.
- share: This value specifying the type of access other threads have to the file.
Exceptions:
- ArgumentException: The path is a zero-length string, contains only white space, or one or more invalid characters as defined by InvalidPathChars. OR access specified Read and mode specified Create, CreateNew, Truncate, or Append.
- ArgumentNullException: The path is null.
- PathTooLongException: The specified path, file name, or both exceed the system-defined maximum length.
- DirectoryNotFoundException: The specified path is invalid.
- IOException: An I/O error occurred while opening the file.
- UnauthorizedAccessException: The path specified a file that is read-only and access is not Read. OR path specified a directory. OR the caller does not have the required permission. OR mode is Create and the specified file is a hidden file.
- ArgumentOutOfRangeException: The mode, access, or share specified an invalid value.
- FileNotFoundException: The file specified in the path was not found.
- NotSupportedException: The path is in an invalid format.
Return Value: Returns a FileStream on the specified path, having the specified mode with read, write, or read/write access and the specified sharing option.
Below are the programs to illustrate the File.Open(String, FileMode, FileAccess, FileShare) method.
Program 1: Below code creates a temporary file, writes some specified contents into it, open that file and print its contents. Here file sharing is not allowed.
CSharp
using System;
using System.IO;
using System.Text;
class GFG {
public static void Main()
{
string path = Path.GetTempFileName();
using (FileStream fs = File.Open(path, FileMode.Open))
{
Byte[] info = new UTF8Encoding( true ).GetBytes( "GFG is a CS Portal." );
fs.Write(info, 0, info.Length);
}
using (FileStream fs = File.Open(path, FileMode.Open,
FileAccess.Read, FileShare.None))
{
byte [] b = new byte [1024];
UTF8Encoding temp = new UTF8Encoding( true );
while (fs.Read(b, 0, b.Length) > 0) {
Console.WriteLine(temp.GetString(b));
}
}
}
}
|
Executing:
GFG is a CS Portal.
Program 2: Initially, a file file.txt is created with some contents shown below:
This code will open the file file.txt and will print its contents with file sharing is not allowed.
CSharp
using System;
using System.IO;
using System.Text;
class GFG {
public static void Main()
{
string path = @"file.txt" ;
using (FileStream fs = File.Open(path, FileMode.Open,
FileAccess.Read, FileShare.None))
{
byte [] b = new byte [1024];
UTF8Encoding temp = new UTF8Encoding( true );
while (fs.Read(b, 0, b.Length) > 0) {
Console.WriteLine(temp.GetString(b));
}
}
}
}
|
Executing:
GeeksforGeeks
Last Updated :
13 Apr, 2021
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