Set environment variables in C - c

Question

Is there a way to set environment variables in Linux using C?
I tried setenv() and putenv(), but they don't seem to be working for me.

Answer 1

I'm going to make a wild guess here, but the normal reason that these functions appear to not work is not because they don't work, but because the user doesn't really understand how environment variables work. For example, if I have this program:
int main(int argc, char **argv)
{
putenv("SomeVariable=SomeValue");
return 0;
}
And then I run it from the shell, it won't modify the shell's environment - there's no way for a child process to do that. That's why the shell commands that modify the environment are builtins, and why you need to source a script that contains variable settings you want to add to your shell, rather than simply running it.

Answer 2

Any unix program runs in a separate process from the process which starts it; this is a 'child' process.
When a program is started up -- be that at the command line or any other way -- the system creates a new process which is (more-or-less) a copy of the parent process. That copy includes the environment variables in the parent process, and this is the mechanism by which the child process 'inherits' the environment variables of its parent. (this is all largely what other answers here have said)
That is, a process only ever sets its own environment variables.
Others have mentioned sourcing a shell script, as a way of setting environment variables in the current process, but if you need to set variables in the current (shell) process programmatically, then there is a slightly indirect way that it's possible.
Consider this:
% cat envs.c
#include <stdio.h>
int main(int argc, char**argv)
{
int i;
for (i=1; i<argc; i++) {
printf("ENV%d=%s\n", i, argv[i]);
}
}
% echo $ENV1
% ./envs one two
ENV1=one
ENV2=two
% eval `./envs one two`
% echo $ENV1
one
%
The built-in eval evaluates its argument as if that argument were typed at the shell prompt. This is a sh-style example; the csh-style variant is left as an exercise!

Answer 3

The environment variable set by setenv()/putenv() will be set for the process that executed these functions and will be inherited by the processes launched by it. However, it will not be broadcasted into the shell that executed your program.
Why isn't my wrapper around setenv() working?

Answer 4

The environment block is process-local, and copied to child processes. So if you change variables, the new value only affects your process and child processes spawned after the change. Assuredly it will not change the shell you launched from.

Answer 5

Not an answer to this question, just wanna say that putenv is dangerous, use setenv instead.
putenv(char *string) is dangerous for the reason that all it does is simply append the address of your key-value pair string to the environ array. Therefore, if we subsequently modify the bytes pointed to by string, the change will affect the process environment.
#include <stdlib.h>
int main(void) {
char new_env[] = "A=A";
putenv(new_env);
// modifying your `new_env` also modifies the environment
// vairable
new_env[0] = 'B';
return EXIT_SUCCESS;
}
Since environ only stores the address of our string argument, string has to be static to prevent the dangling pointer.
#include <stdlib.h>
void foo();
int main(void) {
foo();
return EXIT_SUCCESS;
}
void foo() {
char new_env[] = "A=B";
putenv(new_env);
}
When the stack frame for foo function is deallocated, the bytes of new_env are gone, and the address stored in environ becomes a dangling pointer.