Specifying hex-escaped character followed by a literal hex digit

Question

The following program...

int main () {
    const char str1[] = "\x1bK"; // intent: \x1b followed by 'K'
    const char str2[] = "\x1bE"; // intent: \x1b followed by 'E'
    const char str3[] = "\x1b0"; // intent: \x1b followed by '0'
}

...generates the following compiler warnings for str2 and str3 in all compilers I checked:

warning: hex escape sequence out of range

My intent is to declare three two-character long strings:

str1: \x1b + K
str2: \x1b + E
str3: \x1b + 0

But instead its parsing the last two as if 1be and 1b0 were the specified hex codes.

This seems to happen any time \x?? is followed by a third perceived hexadecimal digit.

In a string literal, how can I specify a character with \x, followed by another character, if that character is also a hexadecimal digit?

I feel like this is a really silly question, but somehow I've never been in this situation before.

That warning is correct: `\x` consumes **all** of the subsequent hex digits. It doesn't know or care how its value is being used. — Pete Becker, Jul 04 '23 at 17:10
@chris Thank you for pointing this out. If I understand correctly [that page](https://en.cppreference.com/w/cpp/language/escape), in C++23 we can also write `"\N{ESC}E"` which might be slightly more readable. — prapin, Jul 04 '23 at 18:47
@prapin, Yes, this is one of those cases where a name would be far clearer, thanks. — chris, Jul 04 '23 at 20:40

score 2 · Accepted Answer · answered Jul 04 '23 at 17:09

2

You could simply have the compiler concatenate 2 separate string literals.

int main () {
    const char str1[] = "\x1b" "K";
    const char str2[] = "\x1b" "E";
    const char str3[] = "\x1b" "0";
}

answered Jul 04 '23 at 17:09

fabian

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The first one doesn't need to do it. – 273K Jul 04 '23 at 17:14
1

@273K -- no, not strictly needed, but it makes the parallel structure clearer. – Pete Becker Jul 04 '23 at 19:51

Specifying hex-escaped character followed by a literal hex digit

1 Answers1