Lines Matching defs:have
18 You should have received a copy of the GNU General Public License
267 size_t have;
271 have = *bufend - *buf;
272 if (have > need)
276 want = need - have;
280 newbuf = (char *) bfd_realloc (*buf, (bfd_size_type) have + want);
284 *bufend = *buf + have + want;
959 /* The two BFD's have the same endianness, and we don't have
990 /* The two BFD's have different endianness, so we must swap
1527 that the information (the pointers and counts) in *DEBUG have been
1779 that have a least one procedure descriptor in them. The final
1804 if (fdr_ptr->cpd == 0) /* Skip FDRs that have no PDRs. */
1942 don't have it already. */
1975 (b) It is also possible for a PDR to have a *lower* vma than its associated
1992 Probably, a better solution would be to have a sorted PDR table. Each
1993 PDR would have a pointer to its FDR so file information could still be
1998 There is still at least one remaining issue. Sometimes a FDR can have a
2020 Since I don't have time to prepare a real fix for this right now, be
2079 file, even though their code may have been generated at
2097 to address ADDR. Thus, after invoking lookup(), we have a
2216 /* If fdr_ptr->rss is -1, then this file does not have full