sbase/expr.c

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/* See LICENSE file for copyright and license details. */
#include <inttypes.h>
#include <stdio.h>
#include <string.h>
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#include "utf.h"
#include "util.h"
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/* token types for lexing/parsing
* single character operators represent themselves */
enum {
VAL = CHAR_MAX + 1, GE, LE, NE
};
struct val {
char *s; /* iff s is NULL, val is an integer */
intmax_t n;
};
static size_t intlen;
static void
enan(struct val v)
{
if (v.s)
enprintf(2, "syntax error: expected integer got `%s'\n", v.s);
}
static void
ezero(intmax_t n)
{
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if (n == 0)
enprintf(2, "division by zero\n");
}
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static char *
valstr(struct val val, char *buf, size_t bufsiz)
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{
if (val.s)
return val.s;
snprintf(buf, bufsiz, "%"PRIdMAX, val.n);
return buf;
}
static int
valcmp(struct val a, struct val b)
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{
char buf1[intlen], buf2[intlen];
char *astr = valstr(a, buf1, sizeof(buf1));
char *bstr = valstr(b, buf2, sizeof(buf2));
if (!a.s && !b.s)
return (a.n > b.n) - (a.n < b.n);
return strcmp(astr, bstr);
}
/* match vstr against BRE vregx (treat both values as strings)
* if there is at least one subexpression \(...\)
* then return the text matched by it \1 (empty string for no match)
* else return number of characters matched (0 for no match)
*/
static struct val
match(struct val vstr, struct val vregx)
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{
regex_t re;
regmatch_t matches[2];
intmax_t d;
char *s, *p, buf1[intlen], buf2[intlen];
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char *str = valstr(vstr, buf1, sizeof(buf1));
char *regx = valstr(vregx, buf2, sizeof(buf2));;
char anchreg[strlen(regx) + 2];
/* expr(1p) "all patterns are anchored to the beginning of the string" */
snprintf(anchreg, sizeof(anchreg), "^%s", regx);
enregcomp(3, &re, anchreg, 0);
if (regexec(&re, str, 2, matches, 0)) {
regfree(&re);
return (struct val){ (re.re_nsub ? "" : NULL), 0 };
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}
if (re.re_nsub) {
regfree(&re);
s = str + matches[1].rm_so;
p = str + matches[1].rm_eo;
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*p = '\0';
d = strtoimax(s, &p, 10);
if (*s && !*p) /* string matched by subexpression is an integer */
return (struct val){ NULL, d };
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/* FIXME? string is never free()d, worth fixing?
* need to allocate as it could be in buf1 instead of vstr.s */
return (struct val){ enstrdup(3, s), 0 };
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}
regfree(&re);
str += matches[0].rm_so;
return (struct val){ NULL, utfnlen(str, matches[0].rm_eo - matches[0].rm_so) };
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}
/* ops points to a stack of operators, opp points to one past the last op
* vals points to a stack of values , valp points to one past the last val
* guaranteed that opp != ops
* ops is unused here, but still included for parity with vals
* pop operator, pop two values, apply operator, push result
*/
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static void
doop(int *ops, int **opp, struct val *vals, struct val **valp)
{
struct val ret, a, b;
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int op;
/* For an operation, we need a valid operator
* and two values on the stack */
if ((*opp)[-1] == '(')
enprintf(2, "syntax error: extra (\n");
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if (*valp - vals < 2)
enprintf(2, "syntax error: missing expression or extra operator\n");
a = (*valp)[-2];
b = (*valp)[-1];
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op = (*opp)[-1];
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switch (op) {
case '|':
if ( a.s && *a.s) ret = (struct val){ a.s , 0 };
else if (!a.s && a.n) ret = (struct val){ NULL, a.n };
else if ( b.s && *b.s) ret = (struct val){ b.s , 0 };
else ret = (struct val){ NULL, b.n };
break;
case '&':
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if (((a.s && *a.s) || a.n) && ((b.s && *b.s) || b.n))
ret = a;
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else
ret = (struct val){ NULL, 0 };
break;
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case '=': ret = (struct val){ NULL, valcmp(a, b) == 0 }; break;
case '>': ret = (struct val){ NULL, valcmp(a, b) > 0 }; break;
case GE : ret = (struct val){ NULL, valcmp(a, b) >= 0 }; break;
case '<': ret = (struct val){ NULL, valcmp(a, b) < 0 }; break;
case LE : ret = (struct val){ NULL, valcmp(a, b) <= 0 }; break;
case NE : ret = (struct val){ NULL, valcmp(a, b) != 0 }; break;
case '+': enan(a); enan(b); ret = (struct val){ NULL, a.n + b.n }; break;
case '-': enan(a); enan(b); ret = (struct val){ NULL, a.n - b.n }; break;
case '*': enan(a); enan(b); ret = (struct val){ NULL, a.n * b.n }; break;
case '/': enan(a); enan(b); ezero(b.n); ret = (struct val){ NULL, a.n / b.n }; break;
case '%': enan(a); enan(b); ezero(b.n); ret = (struct val){ NULL, a.n % b.n }; break;
case ':': ret = match(a, b); break;
}
(*valp)[-2] = ret;
(*opp)--;
(*valp)--;
}
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/* retrn the type of the next token, s
* if it is a value, place the value in v for use by parser
*/
static int
lex(char *s, struct val *v)
{
intmax_t d;
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char *p, *ops = "|&=><+-*/%():";
/* clean integer */
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d = strtoimax(s, &p, 10);
if (*s && !*p) {
*v = (struct val){ NULL, d };
return VAL;
}
/* one-char operand */
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if (*s && !s[1] && strchr(ops, *s))
return *s;
/* two-char operand */
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if (!strcmp(s, ">=")) return GE;
if (!strcmp(s, "<=")) return LE;
if (!strcmp(s, "!=")) return NE;
/* nothing matched, treat as string */
*v = (struct val){ s, 0 };
return VAL;
}
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/* using shunting-yard to convert from infix to rpn
* https://en.wikipedia.org/wiki/Shunting-yard_algorithm
* instead of creating rpn output to evaluate later, evaluate it immediately as
* it is created
* vals is the value stack, valp points to one past last value on the stack
* ops is the operator stack, opp points to one past last op on the stack
*/
static int
parse(char *expr[], int exprlen)
{
struct val vals[exprlen], *valp = vals, v;
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int ops[exprlen], *opp = ops;
int i, type, lasttype = 0;
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char prec[] = { /* precedence of operators */
['|'] = 1,
['&'] = 2,
['='] = 3, ['>'] = 3, [GE] = 3, ['<'] = 3, [LE] = 3, [NE] = 3,
['+'] = 4, ['-'] = 4,
['*'] = 5, ['/'] = 5, ['%'] = 5,
[':'] = 6,
};
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for (i = 0; i < exprlen; i++) {
switch ((type = lex(expr[i], &v))) {
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case VAL:
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*valp++ = v;
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break;
case '(':
*opp++ = '(';
break;
case ')':
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if (lasttype == '(')
enprintf(2, "syntax error: empty ( )\n");
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while (opp > ops && opp[-1] != '(')
doop(ops, &opp, vals, &valp);
if (opp == ops)
enprintf(2, "syntax error: extra )\n");
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opp--;
break;
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default: /* operator */
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if (prec[lasttype])
enprintf(2, "syntax error: extra operator\n");
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while (opp > ops && prec[opp[-1]] >= prec[type])
doop(ops, &opp, vals, &valp);
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*opp++ = type;
break;
}
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lasttype = type;
}
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while (opp > ops)
doop(ops, &opp, vals, &valp);
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if (valp == vals)
enprintf(2, "syntax error: missing expression\n");
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if (--valp != vals)
enprintf(2, "syntax error: extra expression\n");
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if (valp->s)
printf("%s\n", valp->s);
else
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printf("%"PRIdMAX"\n", valp->n);
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return (valp->s && *valp->s) || valp->n;
}
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/* the only way to get usage() is if the user didn't supply -- and expression
* begins with a -
* expr(1p): "... the conforming application must employ the -- construct ...
* if there is any chance the first operand might be a negative integer (or any
* string with a leading minus"
*/
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static void
usage(void)
{
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enprintf(3, "usage: %s [--] expression\n"
"note : the -- is mandatory if expression begins with a -\n", argv0);
}
int
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main(int argc, char *argv[])
{
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intmax_t n = INTMAX_MIN;
/* Get the maximum number of digits (+ sign) */
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for (intlen = (n < 0); n; n /= 10, ++intlen)
;
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ARGBEGIN {
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default:
usage();
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} ARGEND;
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return !parse(argv, argc);
}