...
Of course, we can also use the __stringify() trick to do so, but it is expensive (bigger size, worse performance) to unstringify and get the number again, the expensive atoi() 'works' for the numeric __NR_*, but not work for (__NR_*_base + offset) like __NR_* definitions (used by ARM and MIPS), a simple interpreter is required for such cases and it is more expensive than atoi().
/* not for ARM and MIPS */ static int atoi(const char *s); #define __get_nr(name) __nr_atoi(__stringify(__NR_##name)) #define __nr_atoi(str) (str[0] == '_' ? -1L : ___nr_atoi(str)) #define ___nr_atoi(str) (str[0] == '(' ? -1L : atoi(str))Welcome more discussion or let's simply throw away this direction ;-)
While it will look horrid the it ought to be possible to get the compiler to evaluate the string.
Since "abc"[2] (etc) is converted to a constant (by gcc and clang except at -O0) and you only need to process "n" "nn" "nnn" "(n + m)" (with variable length n and m) then append some spaces and convert the characters back to digits.
So something that starts: #define dig(c) (c < '0' || c > '9' ? 999999 : c - '0') str[0] == '_' ? -1 : str[0] != '(' ? str[1] == ' ' ? dig(str[0]) : str[2] == '1' ? (dig(str[0]) * 10 + dig(str[1]) : Any unexpected character will expand the 99999 and generate an over-large result. I'm not sure how constant the array index need to be. They may well have to be 'integer constant expressions' so cant depend on a previous str[const] value.
I just found a(nother) clang bug: int f(void) { return "a"[2]; } compiles to just a 'return'.
David
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