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2021-11-01leak tests: mark some status tests as passing with SANITIZE=leakLibravatar Ævar Arnfjörð Bjarmason1-0/+1
Mark some tests that match "*status*" as passing when git is compiled with SANITIZE=leak. They'll now be listed as running under the "GIT_TEST_PASSING_SANITIZE_LEAK=true" test mode (the "linux-leaks" CI target). Signed-off-by: Ævar Arnfjörð Bjarmason <avarab@gmail.com> Signed-off-by: Junio C Hamano <gitster@pobox.com>
2011-10-31t7511: avoid use of reserved filename on Windows.Libravatar Pat Thoyts1-1/+1
PRN is a special filename on Windows to send data to the printer. As this is generated during test 3 substitute an alternate prefix to avoid this. Signed-off-by: Pat Thoyts <patthoyts@users.sourceforge.net> Signed-off-by: Junio C Hamano <gitster@pobox.com>
2011-10-26read-cache.c: fix index memory allocationLibravatar René Scharfe1-0/+50
estimate_cache_size() tries to guess how much memory is needed for the in-memory representation of an index file. It does that by using the file size, the number of entries and the difference of the sizes of the on-disk and in-memory structs -- without having to check the length of the name of each entry, which varies for each entry, but their sums are the same no matter the representation. Except there can be a difference. First of all, the size is really calculated by ce_size and ondisk_ce_size based on offsetof(..., name), not sizeof, which can be different. And entries are padded with 1 to 8 NULs at the end (after the variable name) to make their total length a multiple of eight. So in order to allocate enough memory to hold the index, change the delta calculation to be based on offsetof(..., name) and round up to the next multiple of eight. On a 32-bit Linux, this delta was used before: sizeof(struct cache_entry) == 72 sizeof(struct ondisk_cache_entry) == 64 --- 8 The actual difference for an entry with a filename length of one was, however (find the definitions are in cache.h): offsetof(struct cache_entry, name) == 72 offsetof(struct ondisk_cache_entry, name) == 62 ce_size == (72 + 1 + 8) & ~7 == 80 ondisk_ce_size == (62 + 1 + 8) & ~7 == 64 --- 16 So eight bytes less had been allocated for such entries. The new formula yields the correct delta: (72 - 62 + 7) & ~7 == 16 Reported-by: John Hsing <tsyj2007@gmail.com> Signed-off-by: Rene Scharfe <rene.scharfe@lsrfire.ath.cx> Signed-off-by: Junio C Hamano <gitster@pobox.com>