Created
July 19, 2018 17:38
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Bash scoping. Does this program print 1, 2 or 3?
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| #! /bin/bash | |
| x=1 | |
| f() { | |
| echo "$x" | |
| } | |
| g() { | |
| local x=2 | |
| f | |
| } | |
| x=3 | |
| g |
Because bash functions execute in their caller's scope.
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It prints