Bash – Why let a++ Returns Exit Code 1

arithmeticbashksh

Try it:

$ a=0
$ let a++
$ echo $?
1 # Did the world just go mad?
$ echo $a
1 # Yes, it did.
$ let a++
$ echo $?
0 # We have normality.
$ echo $a
2

Contrast with this:

$ b=0
$ let b+=1
$ echo $?
0

And this (from Sirex):

$ c=0
$ let ++c
$ echo $?
0

What is going on here?

$ bash --version
GNU bash, version 4.1.5(1)-release (x86_64-pc-linux-gnu)

Best Answer

From help let:

Exit Status:
If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise..

Since var++ is post-increment, I guess the last argument does evaluate to zero. Subtle...

A perhaps clearer illustration:

$ let x=-1 ; echo x=$x \$?=$?
x=-1 $?=0
$ let x=0 ; echo x=$x \$?=$?
x=0 $?=1
$ let x=1 ; echo x=$x \$?=$?
x=1 $?=0
$ let x=2 ; echo x=$x \$?=$?
x=2 $?=0
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