LanX has asked for the wisdom of the Perl Monks concerning the following question:

Hi

according to perlop#s/PATTERN/REPLACEMENT/ a double eval isn't a double eval but the eval of an interpolation

e    Evaluate the right side as an expression.
ee   Evaluate the right side as a string then eval the result.

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At least that's how I understand the distinction between "evaluate as an expression" and "evaluate as a string"...

Problem is that I can't see this distinction, ee produces just

eval ( eval 'RHS' )

eval do { RHS } ¹

for me and not

eval "RHS".

Or what is meant here, if "evaluate as a string" isn't supposed to mean "string resp. var interpolation", what else?

some snippets as demonstration

  DB<135> $_='x'
 => "x"

  DB<136> $a='$b'; $b="B"
 => "B"

  DB<137> s/x/$a.$a/r            # -> "$a.$a" (interpolation)
 => "\$b.\$b"

  DB<138> s/x/$a.$a/re           # -> eval '$a.$a'
 => "\$b\$b"

  DB<139> s/x/$a.$a/ree          # -> eval "\$b\$b" not eval "\$b.\$b"

Scalar found where operator expected at (eval 106)[(eval 105)[multi_pe
+rl5db.pl:644]:2] line 1, near "$b$b" ...

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but

  DB<140> s/x/$a.'.'.$a/ree      # -> eval "\$b.\$b"
 => "BB"

  DB<141> s/x/$a.'.'.$a/re       # -> eval '$a.\'.\'.$a'
 => "\$b.\$b"

  DB<142> s/x/$a.'.'.$a/r        # -> "$a.'.'.$a"
 => "\$b.'.'.\$b"

  DB<143> eval "\$b.'.'.\$b"     # what perldoc predicts for line 140
 => "B.B"

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Cheers Rolf

(addicted to the Perl Programming Language and ☆☆☆☆ :)

¹) updated: the RHS is precompiled, but eval 'RHS' means runtime compilation

Replies are listed 'Best First'.
Re: perldoc of s///ee wrong or just misleading?
by dave_the_m (Monsignor) on Dec 28, 2014 at 16:03 UTC
    If hypothetically $& was assignable to, then the following are roughly equivalent: 
      s/X/Y/      /X/ && $& = "Y"
  s/X/Y/e     /X/ && $& = Y
  s/X/Y/ee    /X/ && $& = eval Y

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    Dave.

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[d/l]
      I know what you mean and I agree about what is happening, but the perldoc is wrong!

      it should say 

       ee Execute right side as code then eval the
resulting string.

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      NOT

      ee Evaluate the right side as a string then eval the
result.

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      see:

      lanx@lanx-1005HA:~$ perl -MO=Terse -e ' s/(x)/$1.$1/ree '
LISTOP (0x9f99668) leave [1] 
    OP (0x9fa6040) enter 
    COP (0x9f99688) nextstate 
    PMOP (0x9fa2f20) subst 
        LOGOP (0x9fa2e20) substcont 
            UNOP (0x9fa2f00) entereval [256] # --- string eval of resu
+lt
                UNOP (0x9fa2c58) null              
                    LISTOP (0x9fa2be8) scope    # --- precompiled $1.$
+1
                        OP (0x9fa1b28) null [181] 
                        BINOP (0x9fa2c38) concat [3]  
                            UNOP (0x9fa2cb8) null [15] 
                                PADOP (0x9fa2d30) gvsv  GV (0x9fd4c90)
+ *1 
                            UNOP (0x9fa2c18) null [15] 
                                PADOP (0x9fa2c98) gvsv  GV (0x9fd4c90)
+ *1 
-e syntax OK

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      Cheers Rolf

      (addicted to the Perl Programming Language and ☆☆☆☆ :)

[reply]
[d/l]
[select]
      > If hypothetically $& was assignable to, then the following are roughly equivalent:

      maybe better

        s/PAT/STR/     <=>  m/PAT/ && $& = "STR";
  s/PAT/CODE/e   <=>  m/PAT/ && $& = do {CODE};
  s/PAT/CODE/ee  <=>  m/PAT/ && $& = eval do {CODE};
  s/PAT/CODE/eee <=>  m/PAT/ && $& = eval eval do {CODE};
  ... and so on

      [download]

      Cheers Rolf

      (addicted to the Perl Programming Language and ☆☆☆☆ :)

[reply]
[d/l]
Re: perldoc of s///ee wrong or just misleading?
by LanX (Saint) on Dec 28, 2014 at 16:00 UTC
    running B::Terse shows that the RHS is even precompiled and not evaluated at runtime:

    lanx@lanx-1005HA:~$ perl -MO=Terse -e 's/(x)/$1.$1/re'
LISTOP (0x8eb6668) leave [1] 
    OP (0x8ec3040) enter 
    COP (0x8eb6688) nextstate 
    PMOP (0x8ebff20) subst 
        LOGOP (0x8ebff00) substcont 
            UNOP (0x8ebfbe8) null 
                LISTOP (0x8ebfc38) scope 
                    OP (0x8ebeb28) null [181] 
                    BINOP (0x8ebfc18) concat [3] 
                        UNOP (0x8ebfd30) null [15] 
                            PADOP (0x8ebfe20) gvsv  GV (0x8ef1c90) *1 
                        UNOP (0x8ebfc98) null [15] 
                            PADOP (0x8ebfcb8) gvsv  GV (0x8ef1c90) *1

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    That's why this rather ill construction works

    lanx@lanx-1005HA:~$ perl -e 'sub bla { s/x/BEGIN{print "compiletime\n"
+}/e }'
compiletime

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    But still doesn't make the perldocs clearer...

    Cheers Rolf

    (addicted to the Perl Programming Language and ☆☆☆☆ :)

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[d/l]
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Re: perldoc of s///ee wrong or just misleading? (read more)
by Anonymous Monk on Dec 28, 2014 at 20:52 UTC

    Yes the wording is odd, but like a lot of perl docs, if you keep reading you will get all the details, like: A second e modifier will cause the replacement portion to be evaled before being run as a Perl expression.

    I once wrote

    s/regex/STRING /e
    means replace that matched by regex with string
    s/regex/CODEHERE/e
    means replace that matched by regex with result of code; the e in s///e tells the s///ubstitution operator that the s//CODEHERE/e is code and not a string
    s/regex/CODEHERE/e
    means treat CODEHERE string as code
    s/regex/CODEHERE/ee
    means treat CODEHERE string as code, and treat the return value of that code as code
    s/regex/CODEHERE/ee
    means s/regex/eval CODEHERE/e

    One time I even tried to explain this way :) 

    my $string = 'string';
## s/string/replacement/;
$string->substitute( 'pattern', 'replacement' );
## s/string/codehere/e;
$string->substitute( 'pattern', sub { ... } );
## s/string/codehere/ee;
$string->substitute( 'pattern', sub { eval ... } );
## s/string/codehere/eee;
$string->substitute( 'pattern', sub { eval eval ... } );

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[reply]
[d/l]

      Consider 

      s/.../$1' $2"/

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      More accurate: 

      $string->substitute( qr/pattern/, sub { qq/replacement/ } );

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[d/l]
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Re: perldoc of s///ee wrong or just misleading?
by Anonymous Monk on Dec 28, 2014 at 17:16 UTC
    Yes you're right. The wordage is misleading. '/e' and the initial 'e' in '/ee' do the same thing, but the documentation gives an impression that they are somehow different.
[reply]
      I do not like to admit it, but this is one area where I still depend on trial-and-error.
      Bill
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Re: perldoc of s///ee wrong or just misleading?
by ikegami (Patriarch) on Dec 29, 2014 at 17:58 UTC
    SubstitutionExpression evaluated for each matchIf the operator was implemented as a function
    s/pat/repl/qq/repl/subst($_, qr/pat/, sub { qq/repl/ })
    s/pat/repl/ereplsubst($_, qr/pat/, sub { repl })
    s/pat/repl/eeeval(repl)subst($_, qr/pat/, sub { eval(repl) })

    In both of the last two cases, repl is evaluated as a scalar expression and its result is expected to be a string.

    The docs are incorrect or confusing.

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[d/l]
[select]
      > repl is evaluated as a scalar expression

      Not sure, do you mean string eval?

      Cheers Rolf

      (addicted to the Perl Programming Language and ☆☆☆☆ :)

[reply]

        I don't understand the question — neither the whole nor any part of what I said can be replaced by "string eval" and still make sense — but I did say what I meant to say.

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