... or a misunderstanding on my part ?
From string "\x0CTUV", the compiler generates 0x0C 0x55 0x56 0x57 0x00
From string "\x0CABC", the compiler generates 0xCA 0x42 0x43 0x00 ... rather than the expected 0x0C 0x41 0x42 0x43 0x00
I thought the \x escape sequence in a string instructed the compiler to encode the very next two characters as a hexadecimal byte.
Am I missing something ?
The important thing for you is to make sure that the compiler can not consume more characters than you specifically want to belong to the hex constant. My first post shows what I normally do.
I agree. What's odd, however, is why the compiler seemingly ignores leading zeros, rather than consuming all valid hex characters following the \x. Refering back to my original post, you'll see that the "\x0CABC" does NOT consume the B and the C in the string ... both of which by definition are valid hexidecimal digits. It apparently ignores the leading zero, then encodes the next TWO digits ( the C and A in this case ).
Now, given that this forum's primary purpose is a venue for debate rather than assistance, I will accept the obligatory ongoing debate, but you do have to admit that the behavior of the \x conversion is a bit odd. No ?
The compiler may have a fixed number of bits in a character, but that is irrelevant.
Is 0x0 or 0x00 or 0x000 or 0x0000 or 0x00000 or 0x000000 or 0x0000000 or 0x00000000, ... different numbers? All represent the value 0 - a value that fits in a single bit, a char, a short, an int, a long int, a long long int, ...
So it should be obvious that the compiler can not be allowed to count number of digits in a number and stop parsing after a specific number of digits, and assuming that the following characters are part of the next symbol in the source code. There is no difference if we are talking about an integer symbol, or an inlined hexadecimal constant in a text string. The compiler reacts to the characters, not to the number of characters. An integer as a symbol is a non-zero count of digits, followed by a character that is not a digit.
\x will consume all combinations of 0..9, a..f, A..F for any length of characters up to any possible compiler-specific limitation of a string. If the compiler can handle 2000 character long strings, then it can consume a very, very long number of hexadecimal characters.
However, most compilers will issue a warning if the hexadecimal constant overflows the max range for a character. I haven't looked into the standard if it is allowed, but there are compilers who will treat too long hex constants as an error. Note that a character does not have to be 8 bits large. Wide characters are normally 16 bit, but some architectures can have completely different character sizes.
I leave it to C51 users to discuss specific limits for the Keil compiler. The important thing for you is to make sure that the compiler can not consume more characters than you specifically want to belong to the hex constant. My first post shows what I normally do.
The deeper reason for this is that C doesn't assume chars are always 8 bits wide.
But C51 ( the MCU specified in this thread ) DOES specifically state chars are 8 bits wide. Please refer to the table of Data Types on page 95 of Users Guide 09.2001
"Each octal or hexadecimal escape sequence is the longest sequence of characters that can constitute the escape sequence" (my emphasis).
OK, but ... 1) In Kiel C51, what exactly is the \x escape sequence supposed encode? A single character? Multiple characters? 2) In the case of \x, is "the longest sequence of characters that can constitute the escape sequence" all hexadecimal characters ?
"There are specific examples in the C99 standard"
I don't have the C99 standard, but C90 specifically states:
So C51 is behaving exactly as specified!
You just assumed that it accepts two non-zero charactersc :)
It will continue to eat characters until the first non-valid character is found.
gcc would consume every signle character in your second string, and then complain about "hex escape sequence out of range".
but do you think it was an unreasonble assumption?
Yes, it was unreasonable, because it contradicts what the language definition says about \x. There are specific examples in the C99 standard demonstrating exactly this pitfall (6.4.4.4 paragraph 14, 6.4.5 paragraph 7).
> Don't assume that it will encode the next two characters.
Well, apparently I shouldn't, but do you think it was an unreasonble assumption? I don't. I've never read that this escape sequence ignores a leading zero when the 3rd character is a valid hex digit. It makes no sense.
Don't assume that it will encode the next two characters. If the next character after your constant is a valid hexadecimal character, I recommend that you do: "\x0C""ABC" to make sure that you have separated your hexadecimal character from any following characters.
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