What is \xb8\x00\x00\x00.... in C language

**c2tarun** · February 28th, 2012

Hi friends,

I was reading an article and I found this line of code for a C program.

Code:

static char abc[7]="\xb8\x00\x00\x00"
                           "\xff\xe0"

What is meaning of the code in double quotes?

**r-senior** · February 28th, 2012

It's a series of hexadecimal escape sequences. Each '\xNN' represents the character with value NN in hexadecimal.

**Bachstelze** · February 28th, 2012

Whoever wrote this code has no idea what they are doing. There is no guarantee that 0xb8, 0xff and 0xe0 will fit in a char.

**alegomaster** · February 28th, 2012

Originally Posted by Bachstelze

Whoever wrote this code has no idea what they are doing. There is no guarantee that 0xb8, 0xff and 0xe0 will fit in a char.

If it is declared unsigned, shouldn't it work. Also does C declare a char as signed when there is no unsigned or signed prefix in the declaration.

**Bachstelze** · February 28th, 2012

Originally Posted by alegomaster

If it is declared unsigned, shouldn't it work. Also does C declare a char as signed when there is no unsigned or signed prefix in the declaration.

It is implementation-defined whether char actually means signed char or unsigned char.

EDIT: It will work if it is declared unsigned, yes. Here it is not.

**MG&TL** · February 28th, 2012

Originally Posted by Bachstelze

It is implementation-defined whether char actually means signed char or unsigned char.

Would I be right in thinking that passing compiler options would set the default char signed-ness? So...the article could suggest passing the -funsigned-char argument to gcc...but maybe I'm wrong.

**Bachstelze** · February 28th, 2012

Originally Posted by MG&TL

Would I be right in thinking that passing compiler options would set the default char signed-ness? So...the article could suggest passing the -funsigned-char argument to gcc...but maybe I'm wrong.

Yes, you're right. So... maybe, but I think that's quite unlikely.