Why does bash post-increment operator set a nonzero result code?

Question

This caught me by surprise: applying the post-increment operator (++) to a zero value variable results in a non-zero result code. That is, given:

#!/bin/bash
myvar=0
let myvar++
echo "result: $?"

Running that (with bash 5.1.0) results in:

result: 1

Why does that produce a nonzero result code? We see the same behavior using a numeric expression:

#!/bin/bash
myvar=0
(( myvar++ ))
echo "result: $?"

On the other hand, if we use += instead of ++, or if we start with a nonzero value of myvar, we receive a 0 result code as expected. The following...

myvar=1
let myvar+=1
echo "result: $?"

myvar=1
let myvar++
echo "result: $?"

myvar=1
(( myvar++ ))
echo "result: $?"

...all produce:

result: 0

What's going on here?

score 1 · Accepted Answer · edited Jun 02 '21 at 03:44

1

From the bash man page, in the section describing let:

Each arg is an arithmetic expression to be evaluated (see ARITHMETIC EVALUATION). If the last arg evaluates to 0, let returns 1; 0 is returned otherwise

It's not the operator that establishes the value of $?, it is let. Since the value of the argument to let in the command let myvar++ is 0, let returns 1.

edited Jun 02 '21 at 03:44

larsks

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answered Jun 02 '21 at 03:42

William Pursell

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Spot on with the documentation; thanks! That sentence remains unclear to me. In `myvar=0; let myvar++`, I would say that the last arg evaluates to `1`. I would expect `$? == 1` after, say, `myvar=1; let myvar--`. – larsks Jun 02 '21 at 03:50
The operator is called **post**increment, not **pre**increment. You get as a result the value **before** incrementation. Hence, `x=1; echo $((1 + x++))` returns 2, not 3. – user1934428 Jun 02 '21 at 06:59
1

If `myvar` is set to 0, then @larsks `myvar++` evaluates to 0, while `++myvar` evaluates to 1. Both change `myvar` so that its new value is 1. – William Pursell Jun 02 '21 at 11:16

score 1 · Answer 2 · edited Jun 02 '21 at 13:34

For post-increment expressions, bash evaluates the variable and sets the result code before applying the increment. In each case where the value before the post-increment was 0, it follows the documented behavior,

If the last arg evaluates to 0, let returns 1; 0 is returned otherwise.

Note the difference if you use pre-increment,

myvar=0
let ++myvar
echo "result: $?"

>> result: 0

In exp0() in https://git.savannah.gnu.org/cgit/bash.git/tree/expr.c#n1014 , for PREINC it binds the (stringifed) value of v2 then assigns it to val, but under POSTINC it binds the variable then discards v2 without assigning it to val.

Why does bash post-increment operator set a nonzero result code?

2 Answers2