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Appending 0 before binary number based on a value in a variable

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0

I am writing a code where I need to convert some forms of no to binary no for some processing. I want the size of binary no based on the value stored in a variable for which I need to append some leading zeroes.
for example, say there is variable $size =15
Now I want to convert a no from decimal to 15 bit binary

I know how to manually give the count
for ex -
if we want 8 digit binary no from some decimal no then we will use-

 $data_binary = sprintf( "%08b", $initial_data );

But how to specify the size of binary no if the size is stored in some variable??

For example=

 If size =10
 decimal no =12
 required binary no = 0000001100

perl

share|improve this question

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of How do I use sprintf to zero fill to a variable length in Perl?
  
  – Dada
  Mar 26 at 7:44
  

  
  
  
  

add a comment | 

0

I am writing a code where I need to convert some forms of no to binary no for some processing. I want the size of binary no based on the value stored in a variable for which I need to append some leading zeroes.
for example, say there is variable $size =15
Now I want to convert a no from decimal to 15 bit binary

I know how to manually give the count
for ex -
if we want 8 digit binary no from some decimal no then we will use-

 $data_binary = sprintf( "%08b", $initial_data );

But how to specify the size of binary no if the size is stored in some variable??

For example=

 If size =10
 decimal no =12
 required binary no = 0000001100

perl

share|improve this question

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of How do I use sprintf to zero fill to a variable length in Perl?
  
  – Dada
  Mar 26 at 7:44
  

  
  
  
  

add a comment | 

I am writing a code where I need to convert some forms of no to binary no for some processing. I want the size of binary no based on the value stored in a variable for which I need to append some leading zeroes.
for example, say there is variable $size =15
Now I want to convert a no from decimal to 15 bit binary

I know how to manually give the count
for ex -
if we want 8 digit binary no from some decimal no then we will use-

 $data_binary = sprintf( "%08b", $initial_data );

But how to specify the size of binary no if the size is stored in some variable??

For example=

 If size =10
 decimal no =12
 required binary no = 0000001100

perl

share|improve this question

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

 $data_binary = sprintf( "%08b", $initial_data );

 If size =10
 decimal no =12
 required binary no = 0000001100

share|improve this question

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

share|improve this question

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

asked Mar 26 at 7:23

rikkirikki

10999 bronze badges

2 Answers
2

active

oldest

votes

1

See perldoc -f sprintf:

$data_binary = sprintf( "%0*b", $size, $initial_data );

For instance,

printf "%0*bn", 10, 12

prints

0000001100

If the binary representation of your number contains more than $size bits, it won't be truncated. Depending on your use case, you might want to add a substr afterward (but I'd be surprised if you'd actually need to do that).

share|improve this answer

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just found out that this is a duplicate; gonna delete this answer.
  
  – Dada
  Mar 26 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, I definitely don't like the accepted answer, so I'm undeleting this answer, which I think provides a much better way to achieve what OP wants.
  
  – Dada
  Mar 26 at 9:43
  
  

  
  
  
  

add a comment | 

-1

You can use one sprintf to generate format string for another sprintf.

[ There is more than one way to do it ]

use strict; use warnings;
my $binary_length = 10;
my $initial_data = 12;

my $format = sprintf('%%0%db', int($binary_length));
my $data_binary = sprintf( $format, $initial_data );
print $data_binary,"n";

share|improve this answer

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That's unnecessarily complicated. Also, why are you converting $binary_length to a float first?
  
  – melpomene
  Mar 26 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @melpomene 1. It is "general" (any programming language) way. It may be easily improved for perl. 2. int($binary_length) - I prefer such extra safeguards.
  
  – AnFi
  Mar 26 at 8:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  int($binary_length) is not a safeguard. It unnecessarily converts the number to a float.
  
  – melpomene
  Mar 26 at 8:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This solution is not general: It only works in programming languages where printf can take a runtime string. A truly general solution would be sprintf '%0*b', $binary_length, $initial_data, which is also the shortest / simplest solution (except %b is non-standard).
  
  – melpomene
  Mar 26 at 8:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AnFi This question is a duplicate. The better answers on the dup target.
  
  – Dada
  Mar 26 at 9:40
  

  
  
  
  

 | 
show 1 more comment

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1

See perldoc -f sprintf:

$data_binary = sprintf( "%0*b", $size, $initial_data );

For instance,

printf "%0*bn", 10, 12

prints

0000001100

If the binary representation of your number contains more than $size bits, it won't be truncated. Depending on your use case, you might want to add a substr afterward (but I'd be surprised if you'd actually need to do that).

share|improve this answer

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just found out that this is a duplicate; gonna delete this answer.
  
  – Dada
  Mar 26 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, I definitely don't like the accepted answer, so I'm undeleting this answer, which I think provides a much better way to achieve what OP wants.
  
  – Dada
  Mar 26 at 9:43
  
  

  
  
  
  

add a comment | 

1

See perldoc -f sprintf:

$data_binary = sprintf( "%0*b", $size, $initial_data );

For instance,

printf "%0*bn", 10, 12

prints

0000001100

If the binary representation of your number contains more than $size bits, it won't be truncated. Depending on your use case, you might want to add a substr afterward (but I'd be surprised if you'd actually need to do that).

share|improve this answer

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just found out that this is a duplicate; gonna delete this answer.
  
  – Dada
  Mar 26 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, I definitely don't like the accepted answer, so I'm undeleting this answer, which I think provides a much better way to achieve what OP wants.
  
  – Dada
  Mar 26 at 9:43
  
  

  
  
  
  

add a comment | 

See perldoc -f sprintf:

$data_binary = sprintf( "%0*b", $size, $initial_data );

For instance,

printf "%0*bn", 10, 12

prints

0000001100

If the binary representation of your number contains more than $size bits, it won't be truncated. Depending on your use case, you might want to add a substr afterward (but I'd be surprised if you'd actually need to do that).

share|improve this answer

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

$data_binary = sprintf( "%0*b", $size, $initial_data );

printf "%0*bn", 10, 12

0000001100

share|improve this answer

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

answered Mar 26 at 7:37

DadaDada

1,90211 gold badge1010 silver badges2121 bronze badges

-1

You can use one sprintf to generate format string for another sprintf.

[ There is more than one way to do it ]

use strict; use warnings;
my $binary_length = 10;
my $initial_data = 12;

my $format = sprintf('%%0%db', int($binary_length));
my $data_binary = sprintf( $format, $initial_data );
print $data_binary,"n";

share|improve this answer

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That's unnecessarily complicated. Also, why are you converting $binary_length to a float first?
  
  – melpomene
  Mar 26 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @melpomene 1. It is "general" (any programming language) way. It may be easily improved for perl. 2. int($binary_length) - I prefer such extra safeguards.
  
  – AnFi
  Mar 26 at 8:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  int($binary_length) is not a safeguard. It unnecessarily converts the number to a float.
  
  – melpomene
  Mar 26 at 8:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This solution is not general: It only works in programming languages where printf can take a runtime string. A truly general solution would be sprintf '%0*b', $binary_length, $initial_data, which is also the shortest / simplest solution (except %b is non-standard).
  
  – melpomene
  Mar 26 at 8:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AnFi This question is a duplicate. The better answers on the dup target.
  
  – Dada
  Mar 26 at 9:40
  

  
  
  
  

 | 
show 1 more comment

-1

You can use one sprintf to generate format string for another sprintf.

[ There is more than one way to do it ]

use strict; use warnings;
my $binary_length = 10;
my $initial_data = 12;

my $format = sprintf('%%0%db', int($binary_length));
my $data_binary = sprintf( $format, $initial_data );
print $data_binary,"n";

share|improve this answer

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That's unnecessarily complicated. Also, why are you converting $binary_length to a float first?
  
  – melpomene
  Mar 26 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @melpomene 1. It is "general" (any programming language) way. It may be easily improved for perl. 2. int($binary_length) - I prefer such extra safeguards.
  
  – AnFi
  Mar 26 at 8:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  int($binary_length) is not a safeguard. It unnecessarily converts the number to a float.
  
  – melpomene
  Mar 26 at 8:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This solution is not general: It only works in programming languages where printf can take a runtime string. A truly general solution would be sprintf '%0*b', $binary_length, $initial_data, which is also the shortest / simplest solution (except %b is non-standard).
  
  – melpomene
  Mar 26 at 8:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AnFi This question is a duplicate. The better answers on the dup target.
  
  – Dada
  Mar 26 at 9:40
  

  
  
  
  

 | 
show 1 more comment

You can use one sprintf to generate format string for another sprintf.

[ There is more than one way to do it ]

use strict; use warnings;
my $binary_length = 10;
my $initial_data = 12;

my $format = sprintf('%%0%db', int($binary_length));
my $data_binary = sprintf( $format, $initial_data );
print $data_binary,"n";

share|improve this answer

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

use strict; use warnings;
my $binary_length = 10;
my $initial_data = 12;

my $format = sprintf('%%0%db', int($binary_length));
my $data_binary = sprintf( $format, $initial_data );
print $data_binary,"n";

share|improve this answer

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges

answered Mar 26 at 8:02

AnFiAnFi

8,22033 gold badges1717 silver badges4141 bronze badges