将函数局部变量导出到环境

Question

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.

Answer 1

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.

Answer 2

declare inside a functions defaults to local . Use -g to declare a global from inside a function.