You made two changes. You removed the parens around the arguments, and you added parens around substr. The former is inconsequential here, so let's start by reverting it to make things clearer. This means you have 

...   substr( $p, 0, 1 )   = '123' ...
... ( substr( $q, 0, 1 ) ) = '123' ...

[download]

There are two different assignment operators: the scalar assignment operator and the list assignment operator.

To which operator = compiles is determined by the left-hand side (LHS) of the operator. If the LHS look "list-ish", a list assignment operator will be used. Otherwise, a scalar assignment operator will be used.

In the first case, the LHS of the assignment is subtr(...). That's not "list-ish", so a scalar assignment operator is used. A scalar assignment operator in scalar context[1] returns its LHS, so this assignment evaluates the result of substr.

In the second case, the LHS of the assignment is ( ... ). That is "list-ish", so a list assignment operator is used. A list assignment operator in scalar context[1] returns the number of scalars to which its RHS evaluates, so this assignment evaluates to (a fresh scalar with value) 1.

The following is a variant of your code that allows us to check the values returned by the assignments. 

use strict;
use warnings;
use feature 'say';

my $p = 'abc';
my $q = 'abc';

my $ref_p = \scalar(   substr( $p, 0, 1 )   = '123' );
my $ref_q = \scalar( ( substr( $q, 0, 1 ) ) = '123' );

say $$ref_p;  # 123 
say $$ref_q;  # 1

substr( $$ref_p, 0, 1 ) = 'x';
substr( $$ref_q, 0, 1 ) = 'x';

say $$ref_p;  # x23
say $$ref_q;  # x

say $p;       # x23bc
say $q;       # 123bc

[download]

Reference: Mini-Tutorial: Scalar vs List Assignment Operator

  1. substr evaluates its first operand in scalar context.

In reply to Re: Why Perl gets confused here? by ikegami
in thread Why Perl gets confused here? by vr

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