Why one char character appears on 4 bytes in C language

Question

I am coding a C code and I want to compare the hexadecimal value of some character with an other. That is working, but with somme value like � I obtained a value on 4 bytes which create a no sense in my comparaison.

void obs_text(char* start, unsigned char debugLevel, unsigned char depth){
    printf("%c -> 0x%x\n", start[0], start[0]);
}

I expected an output with two hexadecimals characters but the actual output is ? -> 0xffffffef. Please does any one understand what happens ? Thank you for your help.

I am compiling with gcc.

Configured with: --prefix=/Library/Developer/CommandLineTools/usr --with-gxx-include-dir=/Library/Developer/CommandLineTools/SDKs/MacOSX10.14.sdk/usr/include/c++/4.2.1
Apple LLVM version 10.0.1 (clang-1001.0.46.3)
Target: x86_64-apple-darwin18.2.0
Thread model: posix

but I also try on a Debian OS with the same problem

We need more information. Can you provide the code that's calling that function? — S.S. Anne, Apr 12 '19 at 20:24
Note that `start[0]` is promoted to `int` when passed to a variadic function. If the `char` you pass is negative, so will the `int` be, which is then treated by `%x` as an unsigned value. — Weather Vane, Apr 12 '19 at 20:35
Try `printf("%c -> %d\n", start[0], start[0]);` and report the output. — chux - Reinstate Monica, Apr 12 '19 at 21:26

John Cummings · Accepted Answer · 2019-04-12T21:32:43.377

3

Because %x means display as an unsigned integer (4 bytes) as hexadecimal. See http://www.cplusplus.com/reference/cstdio/printf/

As noted there, you could use %hhx (added in C99) to get the behavior you were expecting (see more in this answer).

edited Apr 12 '19 at 21:32

answered Apr 12 '19 at 20:21

John Cummings

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score 1 · Answer 2 · answered Apr 12 '19 at 20:41

1

Use %hhx and cast the argument to unsigned char:

printf( "%c - 0x%hhx", start[0], (unsigned char) start[0] );

%x expects its corresponding argument to have type unsigned int. You need to use the hh length modifier to tell it that you're dealing with an unsigned char value.

answered Apr 12 '19 at 20:41

John Bode

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Note: `(unsigned char)` not needed. Both `(unsigned char) start[0]` and `start[0]` are converted to an `int` before calling `printf()`. `printf()` takes that `int` and then cast to `unsinged char` as part of the `%hhx` specifier. – chux - Reinstate Monica Apr 12 '19 at 21:22

Why one char character appears on 4 bytes in C language

2 Answers2