问题
When same library is linked and used with dlopen, same function (sqrt in this example) has different memory addresses. Can you please explain why it is so? Is there some indirection?
# cat dl-test.c
#include <stdio.h>
#include <stdlib.h>
#include <dlfcn.h>
#include <math.h>
#include <inttypes.h>
int main()
{
void *dl, *dl_sqrt;
dl = dlopen("/lib/x86_64-linux-gnu/libm.so.6", RTLD_LAZY);
if (!dl) {
fprintf(stderr, "%s\n", dlerror());
exit(1);
}
dl_sqrt = dlsym(dl,"sqrt");
if (!dl_sqrt) {
fprintf(stderr, "%s\n", dlerror());
exit(1);
}
printf("Address of sqrt %p\n", (void*) sqrt);
printf("Address of (dl)sqrt %p\n", (void*) dl_sqrt);
return 0;
}
#
# gcc dl-test.c -lm -ldl -o dl-test
# ldd dl-test
linux-vdso.so.1 => (0x00007fff9132b000)
libm.so.6 => /lib/x86_64-linux-gnu/libm.so.6 (0x00007f4caee90000)
libdl.so.2 => /lib/x86_64-linux-gnu/libdl.so.2 (0x00007f4caec8c000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f4cae8c1000)
/lib64/ld-linux-x86-64.so.2 (0x0000559a02daa000)
# ./dl-test
Address of sqrt 0x4006d0
Address of (dl)sqrt 0x7fa7ce6f7250
#
回答1:
Read Drepper's (long) paper How To Write Shared Libraries and study carefully the ELF format. See elf(5), objdump(1), ldd(1), readelf(1), ld-linux(8), dlopen(3), dlsym(3)
Is there some indirection?
Yes, the Procedure Linkage Table, see this.
Notice also the special case of dlopen
with a NULL
file path (to get a handle to your entire program), and the various flags to dlopen
....
来源:https://stackoverflow.com/questions/48109700/why-different-memory-addresses-for-a-function-between-direct-linking-and-dlopen