Re: hex numerical literals in 5.1
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- Subject: Re: hex numerical literals in 5.1
- From: Joshua Jensen <jjensen@...>
- Date: 2005年12月22日 10:04:43 -0700
Todor Totev wrote:
The only thing I care right now is the possibility to enter hexadecimal
numbers in my scripts and give them as parameters to C functions.
And to print them of course but Roberto already made the neccessary
changes for that :-)
Here is an updated llex.c's read_numeral from the LuaPlus for Lua 5.1
alpha that does hex numbers as 0x12345678. I'm not sure if it needs
changes to work against Lua 5.1 beta. I haven't had time to revisit the
code base since the beta came out. Certainly, for 64-bit numbers,
numDigits needs to be increased to 16.
Josh
-------
/* LUA_NUMBER */
static void read_numeral (LexState *ls, SemInfo *seminfo) {
int isReal = 0;
int startsWithZero = ls->current == '0';
while (lex_isdigit(ls->current)) {
save_and_next(ls);
}
if (ls->current == '.') {
save_and_next(ls);
if (ls->current == '.') {
save_and_next(ls);
luaX_lexerror(ls,
"ambiguous syntax (decimal point x string concatenation)",
TK_NUMBER);
}
isReal = 1;
}
if (startsWithZero) {
if (ls->current == 'x') {
/* Process a hex number */
int ch = 0;
int c = 0;
int i = 0;
int numDigits = 8;
next(ls);
do {
ch = tolower(ls->current);
if (lex_isdigit(ch))
c = 16*c + (ch-'0');
else if (ch >= 'a' && ch <= 'f')
c = 16*c + (ch-'a') + 10;
next(ls);
ch = tolower(ls->current);
} while (++i<numDigits && (lex_isdigit(ch) || (ch >= 'a' && ch <=
'f')));
seminfo->r = c;
return;
}
}
while (lex_isdigit(ls->current)) {
save_and_next(ls);
}
if (ls->current == 'e' || ls->current == 'E') {
save_and_next(ls); /* read `E' */
if (ls->current == '+' || ls->current == '-')
save_and_next(ls); /* optional exponent sign */
while (lex_isdigit(ls->current)) {
save_and_next(ls);
}
}
save(ls, '0円');
if (!luaO_str2d(luaZ_buffer(ls->buff), &seminfo->r))
luaX_lexerror(ls, "malformed number", TK_NUMBER);
}