The "e" being absent means "treat the replace expression as a double-quote string literal".

$x = '10+4';
print( "$x" );  # 10+4, not 14
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The "e" being present means "treat the replace expression as a Perl expression".
The Perl expression $1 evaluates to the value of $1, which is the string 10+4.

$x = '10+4';
print( $x );  # 10+4, not 14
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s/.../.../ee is an awful* way of saying s/.../eval .../e.
That means that not only is $1 treated as a Perl expression, so is the value returned by it.

$x = '10+4';
print( eval $x );  # 14
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* — In my opinion, something as dangerous as eval EXPR shouldn't be hidden as an simple little "e".

Update: Added illustrative code.

Thanks. But why a single /e is enough in the first case then?

Because in this cases the evaluated string is "10" + 1, which forces perl to treat "10" as a number. If your evaluated string is "10+1", The whole thing is taken as a string literal.

In one case, you want the replacement expression to be a Perl expression,
In the other, you want the value of $1 to be evaluated.

# s/(10)/$1+1/e
$dollar1 = "10";
$replace = $dollar1 + 1;   # One 'e', so Perl expression.
print $replace;            # 11

# s/(10\+4)/$1/e
$dollar1 = "10+4";
$replace = $dollar1;       # One 'e', so Perl expression.
print $replace;            # 10+4

# s/(10\+4)/$1/ee
$dollar1 = "10+4";
$replace = eval $dollar1;  # Two 'e', so Perl expression + eval.
print $replace;            # 14
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Update: Oops, this was supposed to be a reply to Re^2: /e in regexes.

The /e switch treats the replacement as an expression to be executed. This is not the same as evaluating the contents of $1..$n as executable code, i.e. treating the result of the interpolation as executable code, which needs "double eval" (/ee).

You could re-write

s/(10\+4)/$1/ee;
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as

s/(10\+4)/eval $1/e;
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In the case of s/(10\+4)/$1/e the replacement just resolves as the code line

"10+4";
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which yields a string upon execution.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

Think of it this way:

No /e means the right hand side of the s/// is a double quoted string. One makes the right hand side an expression and the result of evaluating that expression is the substituted value. Two makes the right hand side a string eval and the return from that is the right side.

Your first and second case are the equivalent of "$1" and $1 respectively, which since $1 contains the string value '10+4' works out to the same thing. The last is eval "$1" which actually does what you're expecting from the second.

What confuses a lot of people is that when the replacement is a simple expression like $that, the following are equivalent:

s/$this/$that/;
s/$this/$that/e;
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# these are are equivalent:
s/$this/some text/;
s/$this/"some text"/e;

# and so are these:
s/$this/$that was close/;
s/$this/$that." was close"/e;
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