How to assign a variable to the output value of a program

BeyondSora · 02-07-2011, 12:17 AM

so i wrote myself a very simple hellworld program in c++

...the usual stuff
int main()
{
cout << "helloworld" << endl;
return 0;
}

------------------
how do i assign output value of this program to a variable in shell?

Code:

myvar=$(./a.out); echo $myvar

prints "0" as myvar got assigned to the return value

Code:

./a.out|myvar=$?; echo $myvar

doesnt work either

Dark_Helmet · 02-07-2011, 12:43 AM

Your first attempt should work:

Code:

myvar=$(./a.out); echo $myvar

When you run ./a.out on its own, do you get the expected "helloworld"?

EDIT:
Also, what shell are you running? bash? ksh? tcsh?

BeyondSora · 02-07-2011, 01:13 AM

yes I do get "helloworld"

I am using bash

---------------------
this is so confusing, i made myself the same program using python
and it works......
but just not for c++... this doesnt make sense to me D:

Dark_Helmet · 02-07-2011, 01:46 AM

Well, if it works with your python program, then I'd say you've eliminated bash as a suspect. So it's likely something funky with the c++. For comparison, here's a quickie hello world source, compile, and run:

helloworld.cpp

Code:

#include <iostream>

int main( int argc, char *argv[] )
{
  std::cout << "Hello world!" << std::endl;

  return 0;
}

Plus the compilation and command substitution tests:

Code:

$ g++ -o a.out helloworld.cpp
$ ./a.out
Hello world!
$ myvar=$(./a.out); echo "$myvar - test"
Hello world! - test
$ bash --version
GNU bash, version 4.1.5(1)-release (i486-pc-linux-gnu)
Copyright (C) 2009 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.