Little Endian to Big Endian

Discussion in 'C++' started by invincible, Jun 14, 2005.

  1. invincible

    invincible Guest

    hi I wanted to convert Little Endian to big endian for 2 byte value

    for example

    if hex value of aValue = 0102
    required valu e = 0201

    This can be done

    bValue = ( (aValue << 8) | (aValue << 8))

    But how about if value is bValue = F26B
    I am getting FFF2 instead of 6BF2

    Thanks for help
    invincible, Jun 14, 2005
    #1
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  2. invincible

    Jack Klein Guest

    On Tue, 14 Jun 2005 09:46:35 +0530, "invincible"
    <> wrote in comp.lang.c++:

    First, if you are going to post the same question in multiple
    newsgroups, you should cross-post it, not post it separately to each
    group.

    > hi I wanted to convert Little Endian to big endian for 2 byte value


    Two bytes equals 32 bits on one platform I work on, and 64 bits on
    another that I haven't used for a few years. Do you mean a 16-bit
    value?

    > for example
    >
    > if hex value of aValue = 0102
    > required valu e = 0201
    >
    > This can be done


    Yes, it can.

    > bValue = ( (aValue << 8) | (aValue << 8))


    But not like that!

    > But how about if value is bValue = F26B
    > I am getting FFF2 instead of 6BF2
    >
    > Thanks for help


    Bit-shift and logical operators should not be used on signed integer
    types unless you know exactly what you are doing, and even then only
    rarely.

    Used unsigned types.

    #include <stdio.h>

    unsigned int swap_octets(unsigned int aval)
    {
    return ((aval << 8) | (aval >> 8)) & 0xffff;
    }

    int main()
    {
    unsigned short aval;
    unsigned short bval;

    aval = 0x0201;
    bval = (aval << 8) | (aval << 8);

    printf("Your method, %04X produces %04X\n", aval, bval);

    printf("%04X reverses to %04X\n", aval, swap_octets(aval));

    aval = 0xF26B;
    printf("%04X reverses to %04X\n", aval, swap_octets(aval));

    return 0;
    }

    --
    Jack Klein
    Home: http://JK-Technology.Com
    FAQs for
    comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
    comp.lang.c++ http://www.parashift.com/c -faq-lite/
    alt.comp.lang.learn.c-c++
    http://www.contrib.andrew.cmu.edu/~ajo/docs/FAQ-acllc.html
    Jack Klein, Jun 14, 2005
    #2
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  3. invincible wrote:
    > hi I wanted to convert Little Endian to big endian for 2 byte value
    >
    > for example
    >
    > if hex value of aValue = 0102
    > required valu e = 0201
    >
    > This can be done
    >
    > bValue = ( (aValue << 8) | (aValue << 8))


    If aValue is signed, you need to remove the sign extension.

    bValue = ( (aValue << 8) | ( 0xff & (aValue >> 8) ))

    >
    > But how about if value is bValue = F26B
    > I am getting FFF2 instead of 6BF2
    >
    > Thanks for help
    >
    >
    >
    Gianni Mariani, Jun 14, 2005
    #3
  4. "tuvok" <> wrote in message
    news:d8lp4h$aoc$03$-online.com...
    > "invincible" wrote
    > > hi I wanted to convert Little Endian to big endian for 2 byte value
    > >
    > > for example
    > >
    > > if hex value of aValue = 0102
    > > required valu e = 0201
    > >
    > > This can be done
    > >
    > > bValue = ( (aValue << 8) | (aValue << 8))
    > >
    > > But how about if value is bValue = F26B
    > > I am getting FFF2 instead of 6BF2

    >
    > Try this:
    > bValue = (aValue << 8) | (aValue >> 8);


    Hmmm... I think that would work if you make aValue and
    bValue unsigned. But if they're signed, that right shift
    is going to pour-in a bunch of 1s if the msb of aValue
    is a 1.

    example:

    #include <iostream>
    int main(void)
    {
    register signed int Blat = 0x80000000;
    Blat = Blat >> 21;
    std::cout << std::hex << Blat << std::endl;
    return 0;
    }

    Do you expect that to print 400?
    No, it prints fffffc00.

    BUT, if you make it unsigned:

    #include <iostream>
    int main(void)
    {
    register unsigned int Blat = 0x80000000;
    Blat = Blat >> 21;
    std::cout << std::hex << Blat << std::endl;
    return 0;
    }

    then it prints 400 as expected.


    --
    Cheers,
    Robbie Hatley
    Tustin, CA, USA
    email: lonewolfintj at pacbell dot net
    web: home dot pacbell dot net slant earnur slant



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    Robbie Hatley, Jun 14, 2005
    #4
  5. "invincible" <> wrote in message
    news:d8lln6$46b$...
    > hi I wanted to convert Little Endian to big endian for 2 byte value
    >
    > for example
    >
    > if hex value of aValue = 0102
    > required valu e = 0201
    >
    > This can be done
    >
    > bValue = ( (aValue << 8) | (aValue << 8))
    >
    > But how about if value is bValue = F26B
    > I am getting FFF2 instead of 6BF2
    >
    > Thanks for help


    You're sign-extending. You must be using a C++ implimentation
    which uses 16-bit int. (Or you're using short int.)

    Use unsigned ints to avoid that problem:

    #include <iostream>
    int main(void)
    {
    unsigned int aValue, bValue;
    aValue = 0xA857;
    bValue = ((aValue&0x00FF)<<8) | ((aValue&0xFF00)>>8);
    std::cout << std::hex << bValue << std::endl;
    return 0;
    }

    Prints "57A8".

    --
    Cheers,
    Robbie Hatley
    Tustin, CA, USA
    email: lonewolfintj at pacbell dot net
    web: home dot pacbell dot net slant earnur slant



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    Robbie Hatley, Jun 14, 2005
    #5
  6. "Gianni Mariani" <> wrote in message
    news:...
    > invincible wrote:
    >> hi I wanted to convert Little Endian to big endian for 2 byte value
    >>
    >> for example
    >>
    >> if hex value of aValue = 0102
    >> required valu e = 0201
    >>
    >> This can be done
    >>
    >> bValue = ( (aValue << 8) | (aValue << 8))

    >
    > If aValue is signed, you need to remove the sign extension.
    >
    > bValue = ( (aValue << 8) | ( 0xff & (aValue >> 8) ))
    >


    Be aware that the constant 0xff can have its sign extended when it's
    promoted to 16 bits. So

    0xff & (aValue >> 8)

    can become:

    0xffff & (aValue >> 8)

    which isn't what you intended and doesn't kill the sign extension.

    I always write:

    0x0ff & (aValue >> 8)

    to avoid this problem.

    -Dana
    Dana Cartwright, Jun 14, 2005
    #6
  7. invincible

    Panjandrum Guest

    invincible wrote:
    > hi I wanted to convert Little Endian to big endian for 2 byte value


    Only use the predefined functions htonl, htons, ..., and not your own
    hacks.
    Panjandrum, Jun 14, 2005
    #7
  8. invincible

    red floyd Guest

    Panjandrum wrote:
    > invincible wrote:
    >
    >>hi I wanted to convert Little Endian to big endian for 2 byte value

    >
    >
    > Only use the predefined functions htonl, htons, ..., and not your own
    > hacks.
    >


    Not part of the ISO standard. I think they're POSIX, but they aren't
    guaranteed on all systems.
    red floyd, Jun 14, 2005
    #8
  9. invincible

    Old Wolf Guest

    Dana Cartwright wrote:
    > "Gianni Mariani" <> wrote:
    > >
    > > If aValue is signed, you need to remove the sign extension.
    > >
    > > bValue = ( (aValue << 8) | ( 0xff & (aValue >> 8) ))
    > >

    >
    > Be aware that the constant 0xff can have its sign extended when
    > it's promoted to 16 bits.


    Actually it can't. 0xff is (int) 255, and this must remain
    as 255 when it is promoted to anything. In fact it
    does not get promoted at all in this example.

    > I always write:
    >
    > 0x0ff & (aValue >> 8)
    >
    > to avoid this problem.


    0x0ff is identical to 0xff .
    Old Wolf, Jun 14, 2005
    #9
  10. invincible

    Old Wolf Guest

    Robbie Hatley wrote:
    > "tuvok" wrote:
    > >
    > > Try this:
    > > bValue = (aValue << 8) | (aValue >> 8);

    >
    > Hmmm... I think that would work if you make aValue and
    > bValue unsigned. But if they're signed, that right shift
    > is going to pour-in a bunch of 1s if the msb of aValue
    > is a 1.


    It's implementation-defined as to what gets poured into
    those bits. (But 1-filling is very common.)

    > #include <iostream>
    > int main(void)
    > {
    > register signed int Blat = 0x80000000;


    0x80000000 is an unsigned int (or long) that's outside
    of the range of signed int. So you cause I-D behaviour
    by assigning it to a signed int. (IIRC -- it might be undefined).

    To get well-defined behaviour you could write:
    int Blat = INT_MIN;

    but that won't always get you the representation 0x80000000.

    > Blat = Blat >> 21;
    > std::cout << std::hex << Blat << std::endl;
    > return 0;
    > }
    >
    > Do you expect that to print 400?
    > No, it prints fffffc00.


    On your system.
    Old Wolf, Jun 14, 2005
    #10
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