提交 831e9d9e 编写于 作者: S Szabolcs Nagy 提交者: Rich Felker

regex: simplify the {,} repetition parsing logic

上级 25160f1c
......@@ -708,7 +708,7 @@ static const char *parse_dup_count(const char *s, int *n)
return s;
}
static reg_errcode_t parse_dup(tre_parse_ctx_t *ctx, const char *s)
static const char *parse_dup(const char *s, int ere, int *pmin, int *pmax)
{
int min, max;
......@@ -723,19 +723,13 @@ static reg_errcode_t parse_dup(tre_parse_ctx_t *ctx, const char *s)
max > RE_DUP_MAX ||
min > RE_DUP_MAX ||
min < 0 ||
(!(ctx->cflags & REG_EXTENDED) && *s++ != '\\') ||
(!ere && *s++ != '\\') ||
*s++ != '}'
)
return REG_BADBR;
if (min == 0 && max == 0)
ctx->n = tre_ast_new_literal(ctx->mem, EMPTY, -1, -1);
else
ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
if (!ctx->n)
return REG_ESPACE;
ctx->s = s;
return REG_OK;
return 0;
*pmin = min;
*pmax = max;
return s;
}
static int hexval(unsigned c)
......@@ -988,6 +982,8 @@ static reg_errcode_t tre_parse(tre_parse_ctx_t *ctx)
eg. (+), |*, {2}, but assertions are not treated as empty
so ^* or $? are accepted currently. */
for (;;) {
int min, max;
if (*s!='\\' && *s!='*') {
if (!ere)
break;
......@@ -1007,21 +1003,24 @@ static reg_errcode_t tre_parse(tre_parse_ctx_t *ctx)
sense, note however that the RE_DUP_MAX limit can be
circumvented: (a{255}){255} uses a lot of memory.. */
if (*s=='{') {
err = parse_dup(ctx, s+1);
if (err != REG_OK)
return err;
s = ctx->s;
s = parse_dup(s+1, ere, &min, &max);
if (!s)
return REG_BADBR;
} else {
int min=0, max=-1;
min=0;
max=-1;
if (*s == '+')
min = 1;
if (*s == '?')
max = 1;
s++;
ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
if (!ctx->n)
return REG_ESPACE;
}
if (max == 0)
ctx->n = tre_ast_new_literal(ctx->mem, EMPTY, -1, -1);
else
ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
if (!ctx->n)
return REG_ESPACE;
}
nbranch = tre_ast_new_catenation(ctx->mem, nbranch, ctx->n);
......
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