I have the following:
#!/bin/bash a=0 for d in ./*/ ; do ( cd "$d" ((a++)) echo $a ); done Which goes into each directory in my path, increments a and prints a. However, the output is always 1. Why is that?
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2 Answers
From bash(1):
(list) list is executed in a subshell environment (see COMMAND EXECU‐ TION ENVIRONMENT below). Variable assignments and builtin com‐ mands that affect the shell's environment do not remain in effect after the command completes. The return status is the exit status of list.
By simply removing the parenthesis you have around that block of code, you would have something like this:
#!/bin/bash a=0 for d in ./*/ do ((a++)) echo $a done (also slightly more conventionally formatted)
the result is:
1 2 3 4 5 6 7 
- The one liner quick and working
for d in `seq 1 3`;do echo $d ;doneto not pollute the interactive shell and without$awhich is not needed. If needed:a=0;for d in `seq 1 3`;do echo $a ;((a++));doneTimo– Timo2020-10-29 12:32:09 +00:00Commented Oct 29, 2020 at 12:32 - @Timo How does this relate to the question, though? The question is about some set of files.Håkan Lindqvist– Håkan Lindqvist2020-10-29 12:47:22 +00:00Commented Oct 29, 2020 at 12:47
- 1@Timo Ok, I restored the file pattern from the question as I guess that difference is what caused your confusion.Håkan Lindqvist– Håkan Lindqvist2020-10-30 13:36:38 +00:00Commented Oct 30, 2020 at 13:36
- 1No worries, I got your idea. The question is about traversing a dir but more important about the increment. Your answer is correct.Timo– Timo2020-10-30 14:47:20 +00:00Commented Oct 30, 2020 at 14:47
Because you put the loop body in unnecessary ()'s which makes it execute in a subshell if I recall correctly.
