That's completely wrong.

Precedence indicates where implied parentheses are located. Precedence mere dictates that $c, $c += 1 is short for $c, ($c += 1) rather than ($c, $c) += 1. Precedence is not relevant.

Operand evaluation order dictates which of $c and $c += 1 is evaluated first. Since operand evaluation order is left-to-right for the list/comma operator, $c is evaluated before $c += 1, so you got the order wrong too.

You can see the LTR order in the compiled code: 

>perl -MO=Concise,-exec -e"my @a = ( $c, $c += 1 );"
1  <0> enter
2  <;> nextstate(main 1 -e:1) v:{
3  <0> pushmark s

4  <#> gvsv[*c] s               # $c

5  <#> gvsv[*c] s               # $c + 1
6  <$> const[IV 1] s            #
7  <2> add[t4] sKS/2            #

8  <0> pushmark s
9  <0> padav[@a:1,2] lRM*/LVINTRO
a  <2> aassign[t5] vKS/COMMON
b  <@> leave[1 ref] vKP/REFC
-e syntax OK

[download]

You can see the LTR order in this example: 

my $c = 4;
sub c :lvalue { print("$_[0]\n"); $c } 
my @a = ( c("a"), c("b") += 1 );
print("@a\n");

[download]

outputs 

a
b
5 5

[download]

In reply to Re^2: Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by ikegami
in thread Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by smls

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