Why does "A=3; A=4 echo $A" produce 3 (instead of 4) in Bash?

Question

If I understand correctly, the syntax

Var=<something> command

should run the command after setting Var to "something". Then why does "A=3; A=4 echo $A" produces 3 in my bash?

Shloim · Accepted Answer · 2019-11-05T13:54:11.670

7

Variables in bash are evaluated before execution starts and not during execution, so we have a preprocessing stage for the command:

A=4 echo $A

$A is evaluated to the current value of A and replaces it before the execution to:

A=4 echo 3

and only then it is executed, A changes value to 4, and 3 is printed.

edited Nov 05 '19 at 13:54

answered Nov 05 '19 at 13:15

Shloim

1

TIL. This effect can also be demonstrated by comparing it to `A=3; A=4 eval 'echo $A'`, which prints 4. – L3viathan Nov 05 '19 at 13:23
1

@L3viathan or `A=4 sh -c 'echo $A'` – F. Hauri - Give Up GitHub Nov 05 '19 at 13:25
1

Or when trying to understand why `A=4 ; sh -c "A=5; echo $A"` prints 4 – Shloim Nov 05 '19 at 13:27
2

Further, `A=3; A=4 eval "echo $A"` (note the double quotes) prints `3` – gboffi Nov 05 '19 at 13:27
2

And `B=3 echo $B` prints – gboffi Nov 05 '19 at 13:31

score 0 · Answer 2 · answered Nov 05 '19 at 13:20

0

This is because you do not put a semicolon after the second variable assignment.

Try the following instead:

A=3; A=4; echo $A

answered Nov 05 '19 at 13:20

iwita

2

While technically true, you are describing a different effect than what the OP is looking for. – hymie Nov 05 '19 at 13:37

2 Answers2