converting exponential format number to decimal format number

[Fei Liu]

Hi group, is there a quick way to convert an exponential format number
to decimal format number. For example,

13.534e+10 = 1353400000

I can come up a perl function but it's not perly. Can I get some help
please? Thanks,

 

[John Bokma]

13.534e+10

perl -e "print 13.534e+10"
135340000000

perl -e "my $var = 13.534e+10; print length $var"
12

So at least here (WinXP+ActiveState) Perl does this internally.

 

[usenet]

Hi group, is there a quick way to convert an exponential format number
to decimal format number. For example,

13.534e+10 = 1353400000

How about this:

print 13.534e+10;

#prints 135340000000

 

[Fei Liu]

perl -e "print 13.534e+10"
135340000000

perl -e "my $var = 13.534e+10; print length $var"
12

So at least here (WinXP+ActiveState) Perl does this internally.

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code
where it reads this number from a file and the output needs to be
converted to decimal format for another application (say myapp) to use.
Unfortunately, myapp only understands decimal format number.

 

[J. Gleixner]

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26.

I find it prints 1.3534e+27

Look at: perldoc bigint

 

[usenet]

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code

use Math::BigInt;
my $int = Math::BigInt->new('13.534e+26');
print $int->as_int();

#prints 1353400000000000000000000000

 

[xhoster]

use Math::BigInt;
my $int = Math::BigInt->new('13.534e+26');
print $int->as_int();

Or, if you don't mind there being some 9's way out at the end,

printf "%f", 13.534e26

Xho

 

[usenet]

my $int = Math::BigInt->new('13.534e+26');
print $int->as_int();

Or, if you are just doing the one conversion and don't need to retain
the constructor:

print Math::BigInt->new('13.534e+26')->as_int();

 

[John Bokma]

Thanks for your input, but try 13.534e+26

So, you gave a bad example. Always make your problem description as
complete as possible and provide examples that show your specific problem.
This way people can help you better, and you don't waste a lot of time of
other people.

 

[Dr.Ruud]

Fei Liu schreef:

Hi group, is there a quick way to convert an exponential format
number to decimal format number. For example,

13.534e+10 = 1353400000

I can come up a perl function but it's not perly.

This works in a limited way:

perl -we 'printf "%.0f\n", q/9.64e+21/'
9640000000000000000000

 

[Fei Liu]

complete as possible and provide examples that show your specific problem.
This way people can help you better, and you don't waste a lot of time of
other people.

Your time is appreciated.

 

[Ch Lamprecht]

Or, if you are just doing the one conversion and don't need to retain
the constructor:

print Math::BigInt->new('13.534e+26')->as_int();

as_int returns a Math::BigInt object. We have one already.

print Math::BigInt->new('13.534e+26')



Christoph

 

[Tad McClellan]

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code
where it reads this number from a file and the output needs to be

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Then it is not a number!

It is a string.

converted to decimal format for another application (say myapp) to use.

Perl's regular expressions are handy for working on string data:


$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

 

[John Bokma]

[..]

$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

nify :-D

 

[Fei Liu]

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code
where it reads this number from a file and the output needs to be ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Then it is not a number!

It is a string.

converted to decimal format for another application (say myapp) to use.Perl's regular expressions are handy for working on string data:

$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

This one is neat, I'll use it. Thanks!

 

[Fei Liu]

Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code
where it reads this number from a file and the output needs to be ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Then it is not a number!

It is a string.

converted to decimal format for another application (say myapp) to use.Perl's regular expressions are handy for working on string data:

$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

Hmm, there appears to be a bug in this expression, it's assuming the
number string starts with \., for example
perl -e '$str = '1.2345e+10'; $str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x
($2 - length $1) /ie; print $str;'
12345000000

which is wrong.

 

[Ben Morrow]

Perl's regular expressions are handy for working on string data:


$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

Fails for '1.123456e+5'.

Ben

 

[Tad McClellan]

13.534e+10

Perl's regular expressions are handy for working on string data:


$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

Fails for '1.123456e+5'.

Heck, it fails for simple ol' 1.1e-2, even without searching
through the edges of silliness like you did.

If I did _all_ of his programming for him, I'd have to get his paycheck...

 

[Tad McClellan]

13.534e+10

Perl's regular expressions are handy for working on string data:


$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

Fails for '1.123456e+5'.

Fixed (for that case, anyway) with:

$str =~ s{\A 0* (\d*) \. (\d+) e [+]? (\d+) }
{$1 . (($3 > length $2 ) ?
($2 . '0' x ($3 - length $2)) :
(substr($2,0,$3) . '.' . substr($2,$3)))}xie;


which produces '112345.6', but is getting a bit cumbersome.

^^^^^

bits must be pretty big where you come from.

 

[Tad McClellan]

$str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x ($2 - length $1) /ie;

Hmm, there appears to be a bug in this expression,

It is not "apparent" to me...

it's assuming the
number string starts with \.,

No it isn't.

It is assuming that the *part that needs to be changed* starts with \.

The part before the dot is left as it is.

for example
perl -e '$str = '1.2345e+10'; $str =~ s/\.(\d+)e[+](\d+)/ $1 . '0' x
($2 - length $1) /ie; print $str;'
12345000000

which is wrong.

Errr, what would you have the right answer be then?