Perl：创建二进制数并将其转换为十六进制

Question

I want to create a binary number from the given user input.

Input - Array of number

Output - Binary number

A binary number should be created such that it has one on all the places which has been given as input.

In the given case input is 1, 3, and 7 so my binary no should be 1000101 , so it has 1's on 1, 3 and 7 places from left.

@x = [ 1, 3, 7 ];
$z = 0;
for( $i = 0; $i < 10; $i++ ){
    foreach $elem ( @x ){
        if( $elem == $i ){
            join( "", $z, 1 );
        }
        else{
            join( "", $z, 0 );
       }
   }
}
print "Value of z: $z";

After execution, I am getting the value of z as 0.

I need to convert this binary to hexadecimal.

Is there some function which converts binary to hexadecimal?

Answer 1

[ ] creates an array and returns a reference to that array, so you are assigning a single scalar to (poorly named) @x .

You are also misusing join . Always use use strict; use warnings qw( all ); use strict; use warnings qw( all ); ! It would have caught this error.

Fixed:

my @bits = ( 1, 3, 7 );

my $num = 0;
$num |= 1 << $_ for @bits;
                                           #   76543210
printf("0b%b\n", $num);                    # 0b10001010
printf("0x%X\n", $num);                    # 0x8A

It seems that you want 0b1000101 , so we need to correct the indexes.

my @bits_plus_1 = ( 1, 3, 7 );

my $num = 0;
$num |= 1 << ( $_ - 1 ) for @bits_plus_1;
                                           #   6543210
printf("0b%b\n", $num);                    # 0b1000101
printf("0x%X\n", $num);                    # 0x45

Answer 2

A few problems:

• @x = [ 1, 3, 7 ]; is not an array of three integers. It's an array containing a single array reference. What you want is round brackets, not square brackets: @x = ( 1, 3, 7 );

• The string returned by join is not assigned to $z

---

But even then your code is buggy:

• it appends a bit at the end of $z , not the beginning

• there's a trailing zero that has no business being there.