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Top Forums Shell Programming and Scripting File permission check in SFTP mode Post 302482903 by ravin on Wednesday 22nd of December 2010 11:47:20 PM
Old 12-23-2010
Thanks for the reply..

This is to check if the user doesnt have write permission on a file, then he shouldnt be able to get the file.

anyways will check the above code and provide you an update
 

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DEBUGFS_CREATE_U8(9)					      The debugfs filesystem					      DEBUGFS_CREATE_U8(9)

NAME
debugfs_create_u8 - create a debugfs file that is used to read and write an unsigned 8-bit value SYNOPSIS
struct dentry * debugfs_create_u8(const char * name, umode_t mode, struct dentry * parent, u8 * value); ARGUMENTS
name a pointer to a string containing the name of the file to create. mode the permission that the file should have parent a pointer to the parent dentry for this file. This should be a directory dentry if set. If this parameter is NULL, then the file will be created in the root of the debugfs filesystem. value a pointer to the variable that the file should read to and write from. DESCRIPTION
This function creates a file in debugfs with the given name that contains the value of the variable value. If the mode variable is so set, it can be read from, and written to. This function will return a pointer to a dentry if it succeeds. This pointer must be passed to the debugfs_remove function when the file is to be removed (no automatic cleanup happens if your module is unloaded, you are responsible here.) If an error occurs, NULL will be returned. If debugfs is not enabled in the kernel, the value -ENODEV will be returned. It is not wise to check for this value, but rather, check for NULL or !NULL instead as to eliminate the need for #ifdef in the calling code. COPYRIGHT
Kernel Hackers Manual 3.10 June 2014 DEBUGFS_CREATE_U8(9)
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