
Originally Posted by
bigboab
This is my second attempt.

I hope it is of help. You can't have two or more Executable files with the same 'first part' name, for obvious reasons, the computer would not know which one to run. Remembering you only need to enter the first part of an executable file for it to run. So if you have entered CMD.XXX (used to be command.com) which is the DOS shell then it is up and running. If you try to run it again while it is running you will get an error message.
If this does not help please ignore me. Almost everyone else does.

Sorry Bob, that's simply not true.
If you don't specify where the program is, the "PATH" variable determines which folders the system searches and the order in which they are searched.
If you don't specify the extension the "PATHEXT" variable determines which ones are possible matches, but does not affect the order in which they are checked.
For example, if you create a file called cmd.abc in the C:\Windows\System32 folder, then just type "cmd" the system will find cmd.abc before cmd.exe because it is earlier alphabetically. It will then check "PATHEXT" and will normally reject the file because .abc is not in "PATHEXT", and look for the next file. However, if you add .abc
anywhere in "PATHEXT" the system will select that file as a match, because the order in "PATHEXT" is not important.
Once the system has found a matching file, whether by clicking on the file, typing it's full path or by allowing the system to find it for you, it will try to execute the file by comparing the extension with the contents of the registry. If there is no equivalent class in the registry then you get the error message shown in this problem.
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