I want to call ./prog with arguments from args.txt, but meanwhile, ./prog also reads from file input in.txt
I tried cat args.txt | xargs ./prog < in.txt but it doesn't work.
I understand if the program doesn't need input, cat args.txt | xargs ./prog should work.
But in this case, how should I write it..?
Thank you in advance.
GNU xargs has a -a option to pass in arguments from a file.
xargs -a args.txt ./prog < in.txt
If your xargs doesn't have -a, you could try one of these alternatives:
./prog $(< args.txt) < in.txt
cat args.txt | xargs bash -c './prog "$@" < in.txt' --
The second one is ugly but also more robust when given arguments with whitespace, particularly if you use xargs -0.
./prog $(< args.txt) < in.txt works for me. Could you explain the difference between this and ./prog $(cat args.txt) < in.txt? Does '<' also read file contents to stdout, so that the contents are caught by $()? Isn't '<' an input redirection?
$(<file) is a shorter way of writing $(cat file). Basically no difference. Use cat if you find that more readable.cat file won't work if the contents have special characters like whitespaces
./prog $(cat args.txt)won't work if args.txt have special characters. for example, I might want to pass contents: "arg1 arg2" arg3 as 2 arguments, where the 1st argument is "arg1 arg2" as a whole, the 2nd is arg3