Consider the following code:
#!/usr/bin/bash
t_export() {
declare dummy="Hello"
export dummy
echo dummy: $dummy
echo printenv dummy: $(printenv dummy)
}
t_export
echo dummy: $dummy
echo printenv dummy: $(printenv dummy)
Output:
dummy: Hello
printenv dummy: Hello
dummy:
printenv dummy:
How do you explain this? I thought the environment is always global and therefore the variable dummy would also be visible outside the function.
export doesn't copy values in to the current environment. Instead, it sets the export attribute on a name. When a new processes is started, any variables marked with that attribute are copied (along with their current values) into the environment of the new process.
When t_export returns, the variable dummy goes out of scope, which means it is no longer available to be exported to new processes.
declare inside a functions defaults to local. Use -g to declare a global from inside a function.
export has nothing to do with global variables. It marks a name to be copied into the environment of a new child process. It doesn't immediately copy the variable into the current environment.printenv inside the function.printenv (or more precisely, the process to execute the subshell that starts printenv), the values of any current variables marked as exported are copied into the new process. They don't actually exist in the environment of the current process. (It's a subtle distinction, and one that is hard to confirm directly because the shell doesn't provide a way to examine its own environment, but you are seeing the evidence by dummy no longer existing when you run printenv after t_export returns.)proc filesystem, you can see that an exported variable does not appear in /proc/$$/environ.