I need help with my DOS 6.2! I want to make a program called M1.exe located in a directory to open using the command C:\M1
How can I do this?
I would appreciate for help.

Is this a problem? As written, the program would need to be in the root of c: so that

c:\m1

launches it. If it's elsewhere, you would have to include the full pathname when invoking the command or else have the program's location in your PATH statement

Only two things are infinite, the universe and human stupidity, and I'm not sure about the former.

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Or write a batch file at the root that would dig down into the directory and launch it

i.e:

If M1.exe is located in C:\somedir, then the batch file (m1.bat) should be:

CD somedir
M1.exe

(you could get by with just "M1" in the 2nd line)

Then you could execute the M1.bat file from the root.

So many ways - invoking a program from the command line is pretty much the cornerstone of DOS, so it kind of defies belief that this is a stumbling block.

Perhaps there is more to the question than has been stated

Only two things are infinite, the universe and human stupidity, and I'm not sure about the former.

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"Perhaps there is more to the question than has been stated"

Indeed...

If the directory where M1.exe is located is specified in the path statement, then all is needed at the end of autoexec.bat is:

M1

If not, I'd use

cd\
cd name of directory where M1.exe lives
M1

Or did I misunderstand the question?

Skip

try at the c:\ promot cd\m1\m1