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