Why this perl code about eval not work

Yes.  Generally, for every level of eval, add \ per level (which means \\ for beyond the first).

Anytime you have an eval that doesn't work, the first thing to check on is the escapes and quotes and try debugging if they are what you want for the level of eval nesting you are using.

I don't think you can in this case (not positive) as you have variables at different nesting.  Why do you want to use single quotes?

Let me start even more basic - what are you trying to do with the nested evals?  You can probably do something else that would be simpler and much more efficient.

not need to add a \  you need to escape everythin double

so \ -> \\   $ needs \$  etc.      $var -> \$var
Each needs to be escaped again...:   \$var -> \\\$var
Next level:                                              \\\$var -> \\\\\\\$var
etc.

To rephrase what noci (and I) are saying...

For each nesting of eval that something should be evaluated at, you need to add an escape.  qq(...) is equivalent to double-quoting.

Given the example:

my $code = qq(
   \$str =~ m"$pattern";
   for(\$xx = 1; \$xx <=2; $xx++)
   {
      eval('\$backRef = \$\$xx;');
      push(\@\$backRefArr, \$backRef);
   }
);

eval($code);

print "@$backRRefArr";

Select all Open in new window

$pattern will be evaluated when $code is defined
everything with \x becomes x when $code is defined
everything that had \x is evaluated when the first eval occurs (except the things in the second eval because they are enclosed in single quotes)
the single quoted string is then eval'd by the second eval

Why should it:
I translated your $code to the equivalent $code2...

my $a1 = 'a1_v';
my $varName;

my $code = qq(
      $varName = 'a1';
      print \$\$varName;
);
my $code2 = "
    $varName = 'a1';
    print \$\$varName;
    ";

print "code=$code\n";
print "code2=$code2\n";
#(That should explain the first part... qq is realy the same as "..., if you have an ebcdic system qq( ) is simpler than just "...

# This do not work, either
$varName = 'a1';
print "\$$varName"
#This does what you told it to do print $ and the contents of $varName:   result: $a1    (without line feed)...

Select all Open in new window

You may meant it to do:

$varName = 'a1';
eval( "print \$$varName; " );
print "\n";
eval( "print \"\${$varName}\"; " );
print "\n";
eval( "\$x = \"\${$varName}\"; print \$x;" );
print "\n";

Select all Open in new window

again look at the intermediate results:

$varName = 'a1';
print( "print \$$varName; " );
print "\n";
print( "print \"\${$varName}\"; " );
print "\n";
print( "\$x = \"\${$varName}\"; print \$x;" );
print "\n";

Select all Open in new window

Which requires proper quoting, including the " getting quoted, if needed.

They work at printing nothing?.... as there is quite some missing context. Syntax is correct though.
Anyway what is $pattern...
The pattern isn't matched to anything using =~ ....

Please try the code fragments you wish us to evaluate First as such and describe what you get and what you expect...

Doing something like this:

my $a1 = 'v_1';
my $varName = 'a1';
{ # begin closure
no strict 'refs';
print $$varName, "\n";
} # end closure

Select all Open in new window

does not require an eval.  That is a symbolic ref and works fine.

You are doing "use strict; use warnings;" in your code, right?  If not, you ALWAYS should be as it will catch a lot of errors (and I assume you are which is why I put "no strict 'refs'" since strict disallows symbolic references and you should only allow them where expected).

Your code works fine because it uses symbolic refs.  $$xx evaluates to $1 and then $2.

Apparently I shouldn't do symbolic refs or evals from memory (my code doesn't work as posted and I don't see why).  I'll double-check it later and report a snippet that works...