Why does Shell script create an additional file when only one file is specified

I came across some strange behavior when executing certain Linux commands on different machines. The commands are the ones I provided in the my answer to this question

When I run in Bash and on my Macbook Pro, I get the following results:

$ ls
test.txt    test2.txt   test3.txt   test4.txt

$ (ls *.txt > output) && (sed -i -e 's/^/-v /' output)

$ ls 
output      test.txt    test3.txt
output-e    test2.txt   test4.txt

$ cat output
-v test.txt
-v test2.txt
-v test3.txt
-v test4.txt

$ cat output-e
test.txt
test2.txt
test3.txt
test4.txt

There are two files generated – output and output-e.

Now, output-e just has the name of the *.txt files (this is not expected), but the other one has the proper output with the "-v" in front (which is the expected output).

However, when I run this on a Linux system, it simply just gives me the correct output file. Does anyone know what might be causing this extra file output-e to be generated?

>Solution :

You are using BSD version of sed. The option -i takes an argument that specifies the suffix of the backup file. You specified -e as the suffix, so a backup file output-e is created.