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author | Marek BehĂșn <marek.behun@nic.cz> | 2021-05-20 13:23:55 +0200 |
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committer | Tom Rini <trini@konsulko.com> | 2021-05-24 14:21:30 -0400 |
commit | 46c3e29219e445de150e60e371cfec75f6fee524 (patch) | |
tree | bf659babceadbf369d929618f33cfc5d9c63df28 | |
parent | 6f243e25e6c81a8ab4e98b57fa5b749e0297cac3 (diff) | |
download | u-boot-46c3e29219e445de150e60e371cfec75f6fee524.zip u-boot-46c3e29219e445de150e60e371cfec75f6fee524.tar.gz u-boot-46c3e29219e445de150e60e371cfec75f6fee524.tar.bz2 |
string: make memcpy(), memset(), memcmp() and memmove() visible for LTO
It seems that sometimes (happening on ARM64, for example with
turris_mox_defconfig) GCC, when linking with LTO, changes the symbol
names of some functions, for example lib/string.c's memcpy() function to
memcpy.isra.0.
This is a problem however when GCC for a code such as this:
struct some_struct *info = get_some_struct();
struct some struct tmpinfo;
tmpinfo = *info;
emits a call to memcpy() by builtin behaviour, to copy *info to tmpinfo.
This then results in the following linking error:
.../lz4.c:93: undefined reference to `memcpy'
.../uuid.c:206: more undefined references to `memcpy' follow
GCC's documentation says this about -nodefaultlibs option:
The compiler may generate calls to "memcmp", "memset", "memcpy" and
"memmove". These entries are usually resolved by entries in libc.
These entry points should be supplied through some other mechanism
when this option is specified.
Make these functions visible by using the __used macro to avoid this
error.
Signed-off-by: Marek BehĂșn <marek.behun@nic.cz>
Reviewed-by: Simon Glass <sjg@chromium.org>
-rw-r--r-- | lib/string.c | 9 |
1 files changed, 5 insertions, 4 deletions
diff --git a/lib/string.c b/lib/string.c index a0cff8f..ba176fb 100644 --- a/lib/string.c +++ b/lib/string.c @@ -16,6 +16,7 @@ */ #include <config.h> +#include <linux/compiler.h> #include <linux/types.h> #include <linux/string.h> #include <linux/ctype.h> @@ -513,7 +514,7 @@ char *strswab(const char *s) * * Do not use memset() to access IO space, use memset_io() instead. */ -void * memset(void * s,int c,size_t count) +__used void * memset(void * s,int c,size_t count) { unsigned long *sl = (unsigned long *) s; char *s8; @@ -552,7 +553,7 @@ void * memset(void * s,int c,size_t count) * You should not use this function to access IO space, use memcpy_toio() * or memcpy_fromio() instead. */ -void * memcpy(void *dest, const void *src, size_t count) +__used void * memcpy(void *dest, const void *src, size_t count) { unsigned long *dl = (unsigned long *)dest, *sl = (unsigned long *)src; char *d8, *s8; @@ -586,7 +587,7 @@ void * memcpy(void *dest, const void *src, size_t count) * * Unlike memcpy(), memmove() copes with overlapping areas. */ -void * memmove(void * dest,const void *src,size_t count) +__used void * memmove(void * dest,const void *src,size_t count) { char *tmp, *s; @@ -622,7 +623,7 @@ void * memmove(void * dest,const void *src,size_t count) * @ct: Another area of memory * @count: The size of the area. */ -int memcmp(const void * cs,const void * ct,size_t count) +__used int memcmp(const void * cs,const void * ct,size_t count) { const unsigned char *su1, *su2; int res = 0; |