Thread: range of unsigned integer

  1. #1
    Registered User
    Join Date
    Jul 2007

    range of unsigned integer

    Hello friends,

    I am following a book that states that range of unsigned integers is from 0 to 65535 for 16-bit compilers but this simple program has perplexed me:

       unsigned i, j;
       i = 32768;
       j = 32767;
       printf("%d      %d",i,j);
    The output given by this program is -32768 32767.
    Why is it going to the negative side when the range is from 0 to 65535;
    I am using Borland C++ version 3.1 in Windows XP.
    Thanks a lot.

  2. #2
    Registered User
    Join Date
    Aug 2005
    printf doesnt know the type of value you pass ( how should it, you can pass any type you want ).
    It trusts on the format spacifiers that you pass in the format string.
    You passed %d that stands for a signed integer.

  3. #3
    Deathray Engineer MacGyver's Avatar
    Join Date
    Mar 2007
    First of all, using a 16-bit compiler on a 32-bit OS at the present time is pretty dumb. You'll learn a bunch of stuff that is already outdated currently, but will be less and less useful as 64-bit systems are being rolled out.

    Secondly, signed and unsigned is really more of a concept. You're not making the variable itself signed or unsigned. You're giving a hint to the compiler how to actually treat math operations and such. Consider binary representation for a second.

    If I gave the 16-bit number 0xFFFF, what number is it? The signed representation could be -1 while the unsigned representation could be 65535. printf() doesn't have the luxury of knowing what type of number you're giving it because it's a variadic function. It acts on the arguments it receives based upon what is in the first argument. This is why you're using things like %d. %d tells printf() to treat a variable as if it's an integer. The issue is that printf() will treat the value matching %d as if it's signed.

    Hint: Try %u.

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