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comparing signed and unsigned variables

Paul Connor

In the following code:-

char a = 0x80;
unsigned char b = 0x80;

if(a==0x80)
        printf("a==0x80\n");

if(b==0x80)
	printf("b==0x80\n");
The first printf isn't executed but the second is. Why is this? What is the correct way to compare signed variables?
I did think that this might be because the constant was an integer but a=='\x80' produces the same result.

TIA

Paul

Parents
  • 0 in reply to Jon Ward

    Jon,

    I understand what you're saying but if the "0x80" in

    char a = 0x80;
    is actually a -1 then what value is the "0x80" in
    (a==0x80)
    and what about the '\x80' in
    (a=='\x80')
    I instinctively thought that in binary:
    a = 1000 0000
    0x80 = 1000 0000
    '\x80' = 1000 0000

    I supose this is really an academic question, as I know I was wrong and I have fixed the code I found the error in. I'd just like to understand why I was wrong.

    cheers

    Paul

Reply
  • 0 in reply to Jon Ward

    Jon,

    I understand what you're saying but if the "0x80" in

    char a = 0x80;
    is actually a -1 then what value is the "0x80" in
    (a==0x80)
    and what about the '\x80' in
    (a=='\x80')
    I instinctively thought that in binary:
    a = 1000 0000
    0x80 = 1000 0000
    '\x80' = 1000 0000

    I supose this is really an academic question, as I know I was wrong and I have fixed the code I found the error in. I'd just like to understand why I was wrong.

    cheers

    Paul

Children
  • 0 in reply to Paul Connor

    have you tried disabling the ANSI Integer Promotions?

  • 0 in reply to Paul Connor

    what value is the "0x80" in

    (a==0x80)

    0x0080

    and what about the '\x80' in

    (a=='\x80')

    again, 0x0080


    These things are ALL integers as per ANSI. 0x80 is 128 or 0x0080. '\x80' is also 128 or 0x0080.

    The "character" 'A' is 0x0041 and its type is int (not char).

    If you cast the 0x80 to a char, for example:

    (a==(char)0x80)

    you will get the results you want.

    Jon

  • 0 in reply to Jon Ward

    Incidentally, 0x80 in 8-bit twos-complement represents -128, not -1. -1 is all ones, 0xff. (To convert, invert all the bits and add 1.)

    The problem here is in the comparison. In the sequence:

    a = 0x80; // same as a = -128;
    if (a == 0x80)

    the 0x80 is a literal constant of type "int", while a is of type signed char. To compare the two, the compiler will convert the signed char to an int.

    You can represent positive 128 decimal (0x80) in a 16-bit integer, so that's the right side of the comparison. The left side, a, has the value -128. -128 != 128, so the comparison fails.

    Change the test to either
    if (a == (signed char)0x80)
    or
    if (a == -128)
    and see what happens.

    Disabling integer promotion as Andy suggests should make the problem go away as well. (Just remember that integer promotion is standard ANSI behavior.)

    You might also run into this problem when it comes to procedure parameters, particularly with printf(). Read up on the Keil extensions to printf format specifiers, in particular the ones like %bu (print unsigned byte) instead of just %u.

  • 0 in reply to Drew Davis

    Thanks all,

    I understand a lot better now.

  • 0 in reply to Drew Davis

    Incidentally, 0x80 in 8-bit twos-complement represents -128, not -1. -1 is all ones, 0xff.

    Errr. That's what I meant. -128, yeah. -128.

    Jon