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

Hi friends,

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

Code:

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

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What is meaning of the code in double quotes?

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

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.

Quote:

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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.

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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.

Quote:

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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.

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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.

Quote:

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Originally Posted by Bachstelze

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

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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.

Quote:

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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.

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