Running exe with parameters in bash and passing it's output to another program

Question

I have to call one prog1.exe program and give arguments to it and give it's output to another executable prog2.exe in a bash script

prog1 "arguments" | prog2
echo finished

How can I make this to work? Also I suppose echo command won't be executed before prog2 will finish it's work?

EDIT : When I run

prog1 "argument"
prog2 "example"
prog1 "argument" | prog2

I got correct output from prog1 - std::cout << "arguments" << std::endl and it's displayed nicely in console and also prog2 is creating file example.txt so individually both programs are working. Just puting them together with prog1 "arguments" | prog2 doesn't do the trick

You wrote it correctly. Have you tried it, does it not work? — John Kugelman, Aug 31 '20 at 15:21
The `echo` will only run after both `prog1` and `prog2` have exited. — John Kugelman, Aug 31 '20 at 15:22
Yes it's not working. I made prog2 to create text file with name that prog1 gave it but it acts as no parameters are given — etrusks, Aug 31 '20 at 15:23
@etrusks pipes do not add parameters, they send to stdin of the process. — jordanm, Aug 31 '20 at 15:56

score 1 · Accepted Answer · answered Aug 31 '20 at 16:32

This really isn't a bash question. This is a C++ question really. prog2 should be reading from stdin as jordanm suggested in the comments. Something like this will read from stdin line by line:

for (std::string line; std::getline(std::cin, line);) {
        std::cout << line << std::endl;
}

Change the cout to do whatever you want with it on each line.

Running exe with parameters in bash and passing it's output to another program

1 Answers1