I found this code snippet on SO (sorry I don't have the link to the question/answer combo)
bool isDir = (File.GetAttributes(source) & FileAttributes.Directory) == FileAttributes.Directory;
This confuses me because FileAttributes.Directory
is on both sides of the ==
.
What does the &
do in this case? I'm not sure how to read this line of code. I'm trying to evaluate whether a path string is a file or a directory.
It is using a bit mask to test if a single bit (FileAttributes.Directory) is set.
The values of the enum are powers of two, corresponding to individual bits.
ReadOnly = 1,
Hidden = 2,
System = 4,
Directory = 16,
Archive = 32,
Device = 64,
If ReadOnly and Directory are set then FileAttributes is equal to 17. The calculation looks like this in binary:
File.GetAttributes(source) = 00001001
FileAttributes.Directory = 00001000 &
-------------------------------------
00001000
If the Directory bit was not set you'd get zero instead:
File.GetAttributes(source) = 00000001
FileAttributes.Directory = 00001000 &
-------------------------------------
00000000
A slightly more concise way to write the expression that gives the same effect is to test against zero:
bool isDir = (File.GetAttributes(source) & FileAttributes.Directory) != 0;
its doing a Bitwise AND operation. Attributes are stored as bit flags, so it is and'ing those flags together with AttributeFlags.Directory to see if one of the attributes is .Directory.
Good example of Bit Flags here: http://weblogs.asp.net/wim/archive/2004/04/07/109095.aspx
[Flags]
public enum FileAttributes
{
Archive, // 0000
Compressed, // 0001
Device, // 0010
Directory, // 0100
Encrypted, // 1000
...
}
Then:
File.GetAttributes(source): 1101
FileAttributes.Directory: 0100
(Logical AND): 0100
0100 is the same as the directory flag, so we now know that that flag is in the chosen flags of the enum.
It is the logical & operator. In this particular example it checks if the FileAttributes enumeration has the Directory value, verifying if the string pointed by the source
variable is a directory.
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