Different behavior between eval "07" and eval "08"

[Liang Wang]

eval "07" gets 7 but eval "08" gets an undefined value. I tested perl
5.8.8 on fedora 8 and ubuntu.

Is there any difference between those two eval statements? Is this a
bug?

 

[John W. Krahn]

eval "07" gets 7 but eval "08" gets an undefined value. I tested perl
5.8.8 on fedora 8 and ubuntu.

Is there any difference between those two eval statements? Is this a
bug?

No, there is no difference in the eval statements. The problem is that
"08" is not a valid octal number.


John

 

[Jürgen Exner]

eval "07" gets 7 but eval "08" gets an undefined value. I tested perl
5.8.8 on fedora 8 and ubuntu.

Is there any difference between those two eval statements?

Yes. The digit 7 is a valid octal digit, 8 is not a valid octal digit.

jue

 

[Liang Wang]

No, there is no difference in the eval statements. The problem is that
"08" is not a valid octal number.

John

Got it. Thanks.

I've tested "06", but not "09". What I really want to do is to
transform string "08" to integer 8. But in this case, eval doesn't
treat "08" as string, as you said.

 

[Ben Morrow]

eval "07" gets 7 but eval "08" gets an undefined value. I tested perl
5.8.8 on fedora 8 and ubuntu.

Is there any difference between those two eval statements? Is this a
bug?

Nope. Both 07 and 08 *as literals* (i.e. either specified directly in
your source or passed to eval STRING) are interpreted as octal numbers.
07 is a valid octal representation of the number 7; this number can also
be written 0x7 if you like hex, or 0b111 if you like binary . 08 is
not a valid octal representation of any number, since 8 is not an octal
digit.

Why are you trying to eval a string consisting of a single number
anyway? You do know Perl will autoconvert a string to a number if you
use it as one, and in *this* case it always uses decimal? So "08" + 1
evaluates to 9, as you'd expect. If you're trying to 'normalise' the
representation of the number (get rid of the leading zeros, etc.) you
can force a numeric conversion by adding 0.

~% perl -le'print 0 + "08"'
9

Ben

 

[John W. Krahn]

Got it. Thanks.

I've tested "06", but not "09". What I really want to do is to
transform string "08" to integer 8. But in this case, eval doesn't
treat "08" as string, as you said.

The string "08" is already the integer 8.

$ perl -le' $x = "08"; print $x; print $x + 1'
08
9



John

 

[Tad J McClellan]

What I really want to do is to
transform string "08" to integer 8.

Why do you think you need to transform string "08" to integer 8?

Values in Perl are both strings and numbers simultaneously.

You really should review the "Scalar values" and
"Scalar value constructors" sections in perldata.pod.

 

[John W. Krahn]

Nope. Both 07 and 08 *as literals* (i.e. either specified directly in
your source or passed to eval STRING) are interpreted as octal numbers.
07 is a valid octal representation of the number 7; this number can also
be written 0x7 if you like hex, or 0b111 if you like binary . 08 is
not a valid octal representation of any number, since 8 is not an octal
digit.

Why are you trying to eval a string consisting of a single number
anyway? You do know Perl will autoconvert a string to a number if you
use it as one, and in *this* case it always uses decimal? So "08" + 1
evaluates to 9, as you'd expect. If you're trying to 'normalise' the
representation of the number (get rid of the leading zeros, etc.) you
can force a numeric conversion by adding 0.

~% perl -le'print 0 + "08"'
9

I think your version of Perl has a bug in it. Mine produces a different
result.

$ perl -le'print 0 + "08"'
8



John

 

[Ben Morrow]

I think your version of Perl has a bug in it. Mine produces a different
result.

$ perl -le'print 0 + "08"'
8

Heh . Yes, sorry.

Ben