input-file Specifies the .EXE file to be converted.
output-file Specifies the binary file to be created.
...so you might use a command like:
exe2bin D:\applications\thefile.exe D:\bin_store\thenewfile.bin
You do not need to use the drive letter if cmd prompt is in the right drive.
You don't need a path1 if cmd prompt is in the folder containing thefile.exe:
If you do not give a drive and\or path2 then thenewfile.bin will be created alongside thefile.exe.
Of course, you can call the new bin file thefile.bin also.... no clash.
So instead you might go:
cd /d D:\applications
exe2bin thefile.exe thefile.bin
And that is it.
Just mind spaces in path and file names. If the folder was My Applications then you would write:
cd /d d:\"my applications" ...or:
exe2bin D:\"my applications\thefile.exe D:\bin_store\thenewfile.bin"
exe2bin "D:\my applications\thefile.exe" D:\bin_store\thenewfile.bin
...whatever, just so the name with a space is inside quotations.
The .exe file is in the C drive.
When I tried it the error came that "The file cannot be converted".
Can you please send me the code to do so when the file is in C drive.
The path of the .exe file is:- "C:\speed.exe"
the path of the bin could be anywhere in the C drive.
exe2bin is something dragged over into cmd.exe from the DOS era. It is limited to programs of less than 64KB. M$ gives these as the cause of that error:
Cause: The program to be converted has one of the following problems:
* The program has an origin of 0100h but a different entry point.
* The program requires segment fixups.
* The program code and data are larger than 64 KB.
* The program has more than one declared segment
* The file is not a valid .EXE-format file.
exe2bin has found your pgm from the path given, otherwise you would have received a "path not found" error.
exe2bin c:\speed.exe c:\speed.bin