PathCompletionChar Entry
Applies To: Windows Server 2003, Windows Server 2003 R2, Windows Server 2003 with SP1, Windows Server 2003 with SP2
PathCompletionChar
HKCU\Software\Microsoft\Command Processor
| Data type | Range | Default value |
|---|---|---|
REG_DWORD |
0x0 | 0x1–0x1F | 0x20 |
0x9 (the TAB key) |
Description
Specifies a character that activates the directory name completion feature in Command Processor (Cmd.exe).
To specify a character, enter the hexadecimal value of an ASCII control character other than 0x20 (SPACEBAR). For example, 0x4 represents Ctrl+D, and 0x6 represents Ctrl+F.
| Value | Meaning |
|---|---|
0x0 |
Disables completion. |
0x1–0x1F |
Corresponding ASCII control character. |
0xD |
Disables completion. |
0x9 |
TAB key. |
0x20 |
SPACEBAR. Disables completion. |
Change Method
You can change the value of this entry by using Cmd.exe, but only for the duration of that Cmd.exe's execution. For all other changes, you must use the registry editor Regedit.exe.
Activation Method
If you make changes to the value of this entry by using Regedit.exe, you must restart Cmd.exe to make the changes effective.
Notes
PathCompletionChar can appear in HKLM and HKCU. If both appear in the registry, then the value of the entry in HKCU takes precedence over the value in HKLM.
Assigning the values of 0x0, 0xD, or 0x20 to this entry disables file name completion rather than assigning a completion character.
The command-line perameters take precedence over the registry settings while using Cmd.exe.
The datatype of this entry can be either REG_DWORD or REG_SZ.
If you use the console Paste function to enter text into a CMD window, then you might trigger completion if the characters you paste contain completion characters.
Tip
To enable or disable directory name completion for one use of Cmd.exe, at the command prompt, type CMD /F:ON (to enable) or CMD /F:OFF (to disable). This command sets the completion character to CTRL+D (0x4).
For more information about the path completion feature, at the command prompt, type CMD /?.