Set environment variables in C
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.
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 twoENV1=oneENV2=two% eval `./envs one two`% echo $ENV1one% 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!
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.