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regex: simplify the {,} repetition parsing logic
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25160f1c08
commit
831e9d9efa
@ -708,7 +708,7 @@ static const char *parse_dup_count(const char *s, int *n)
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return s;
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}
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static reg_errcode_t parse_dup(tre_parse_ctx_t *ctx, const char *s)
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static const char *parse_dup(const char *s, int ere, int *pmin, int *pmax)
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{
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int min, max;
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@ -723,19 +723,13 @@ static reg_errcode_t parse_dup(tre_parse_ctx_t *ctx, const char *s)
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max > RE_DUP_MAX ||
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min > RE_DUP_MAX ||
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min < 0 ||
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(!(ctx->cflags & REG_EXTENDED) && *s++ != '\\') ||
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(!ere && *s++ != '\\') ||
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*s++ != '}'
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)
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return REG_BADBR;
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if (min == 0 && max == 0)
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ctx->n = tre_ast_new_literal(ctx->mem, EMPTY, -1, -1);
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else
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ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
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if (!ctx->n)
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return REG_ESPACE;
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ctx->s = s;
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return REG_OK;
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return 0;
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*pmin = min;
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*pmax = max;
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return s;
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}
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static int hexval(unsigned c)
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@ -988,6 +982,8 @@ static reg_errcode_t tre_parse(tre_parse_ctx_t *ctx)
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eg. (+), |*, {2}, but assertions are not treated as empty
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so ^* or $? are accepted currently. */
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for (;;) {
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int min, max;
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if (*s!='\\' && *s!='*') {
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if (!ere)
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break;
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@ -1007,21 +1003,24 @@ static reg_errcode_t tre_parse(tre_parse_ctx_t *ctx)
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sense, note however that the RE_DUP_MAX limit can be
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circumvented: (a{255}){255} uses a lot of memory.. */
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if (*s=='{') {
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err = parse_dup(ctx, s+1);
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if (err != REG_OK)
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return err;
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s = ctx->s;
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s = parse_dup(s+1, ere, &min, &max);
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if (!s)
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return REG_BADBR;
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} else {
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int min=0, max=-1;
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min=0;
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max=-1;
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if (*s == '+')
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min = 1;
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if (*s == '?')
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max = 1;
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s++;
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ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
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if (!ctx->n)
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return REG_ESPACE;
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}
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if (max == 0)
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ctx->n = tre_ast_new_literal(ctx->mem, EMPTY, -1, -1);
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else
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ctx->n = tre_ast_new_iter(ctx->mem, ctx->n, min, max, 0);
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if (!ctx->n)
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return REG_ESPACE;
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}
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nbranch = tre_ast_new_catenation(ctx->mem, nbranch, ctx->n);
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