in reply to Decmal to 16 bit binary number

There are two ways to pack the bytes in a 16-bit integer:

(Wikipedia The diagrams on the right part way down are clearest explanation to me.)

Update: Switched the descriptions around per proceng++ post below.

  1. Big-endian.

    Here, the bits of the low-value byte are stored in the byte of the target location with the highest address

    You can do this is perl using my $BEbin = pack 'n', 1263;

  2. Little-endian

    Here, the bits of the low-value byte are stored in the byte of the target location with the lowest address.

    You can do this is perl using my $LEbin = pack 'v', 1263;

There are also two ways of unpacking the bits of either representation:

  1. Least significant bit (lsb) first, progressing to most significant bit (msb).

    print unpack 'b16', $bin

  2. msb first progressing to lsb.

    print npack 'B16', $bin;

The result is 4 different ways of displaying the binary representation of a 16-bit number. By the example you gave, it would appear to be the last of those below that you are after: 

print unpack 'B16', pack 'v', 1263;;
1110111100000100

print unpack 'B16', pack 'n', 1263;;
0000010011101111

print unpack 'b16', pack 'v', 1263;;
1111011100100000

print unpack 'b16', pack 'n', 1263;;
0010000011110111

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Re^2: Decmal to 16 bit binary number
by proceng (Scribe) on Nov 01, 2007 at 13:32 UTC
    1. Big-endian. Here, the bits of the low-value byte are stored in the byte of the target location with the lowest address. You can do this is perl using my $BEbin = pack 'n', 1263;
    2. Little-endian Here, the bits of the low-value byte are stored in the byte of the target location with the highest address You can do this is perl using my $LEbin = pack 'v', 1263;
    Shouldn't this be the other way around?
    In Big-endian, the bytes are stored in memory in order eg: 0x1263 would be 0x1263.
    In Little-endian, 0x1263 would be stored as 0x6312.

    Update: Edited for formatting

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