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Lefteris
Lefteris

Reputation: 3256

Comparing chars with hex values

For the last day I have had a nasty bug in my code which after some searching appears to be related to comparison between char values and hex. My compiler is gcc 4.4.1 running on Windows. I replicated the problem in the simple code below:

char c1 = 0xFF; char c2 = 0xFE;
if(c1 == 0xFF && c2 == 0xFE)
{
    //do something
}

Surprisingly the code above does not get in the loop. I have absolutely no idea why and would really appreciate some help on this. It is so absurd that the solution must be (as always) a huge mistake on my part that I totally overlooked.

If I replace the above with unsigned chars it works, but only in some cases. I am struggling to find out what's going on. In addition if I cast the hex values to char in comparison it enters the loop correctly like so:

if(c1 == (char)0xFF && c2 == (char)0xFE)
{
    //do something
}

What does that mean? Why can it be happening? Isn't the raw hex value interpreted as a char by default? For the curious the point in my code where I first noticed it is comparison of first 2 bytes of a stream with the above hex value and their reverse to idenity the Byte Order Mark.

Any help is appreciated

Upvotes: 13

Views: 33794

Answers (5)

ericcurtin
ericcurtin

Reputation: 1717

When comparing char to hex you must be careful:

Using the == operator to compare a char to 0x80 always results in false?

I would recommend this syntax introduced in C99 to be sure

if(c1 == '\xFF' && c2 == '\xFE')
{
    // do something
}

Avoid the cast, it's unnecessary and isn't type safe.

It tells the compiler that the 0xFF is a char rather than an int, this will solve your issue.

clang compiler will warn you about this also:

comparison of constant 128 with expression of type 'char' is always false [-Werror,-Wtautological-constant-out-of-range-compare]

Upvotes: 9

siggyt
siggyt

Reputation: 9

I solved it by casting my variables to UINT16 (optimized for my compiler). In your case, you would be casting c1 and c2 as INT

char c1 = 0xFF; char c2 = 0xFE;
if((int)c1 == 0xFF && (int)c2 == 0xFE)
{
    //do something
}

Upvotes: 0

paxdiablo
paxdiablo

Reputation: 881373

The literal 0xff is not a char, it's a int (signed). When you shoehorn that into a char variable, it will fit okay but, depending on whether your char type is signed or unsigned, this will affect how it's upgraded in expressions (see below).

In an expression like if (c1 == 0xff), the c1 variable will be promoted to an integer since that's what 0xff is. And what it's promoted to depends on whether it's signed or not.

Bottom line, you can do one of two things.

  1. Ensure that you use signed char so that it "sign-extends" to the correct int. By that I mean an unsigned char 0xff would become (for a 4-byte int) 0x000000ff (so it's still 255) but a signed one would become 0xffffffff (so it's still -1).

  2. Shoehorn the literal into the same type as the variable, which you're already doing with (char)oxff.

Upvotes: 4

Ed Heal
Ed Heal

Reputation: 59997

The character 0xFE will he translated into a negative integer. The constants in the expression will be translated into positive integers.

Upvotes: 0

Jonathan Leffler
Jonathan Leffler

Reputation: 753675

Plain char can be signed or unsigned. If the type is unsigned, then all works as you'd expect. If the type is signed, then assigning 0xFF to c1 means that the value will be promoted to -1 when the comparison is executed, but the 0xFF is a regular positive integer, so the comparison of -1 == 0xFF fails.

Note that the types char, signed char and unsigned char are distinct, but two of them have the same representation (and one of the two is char).

Upvotes: 13

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