Linux程序动态库加载优化

浏览数：41 / 时间：2015年06月20日

  作者：zhanhailiang 日期：2014-10-26

linux程序动态库加载流程简介

linux从程序（program或对象）变成进程（process或进程），简单说来需要经过三步：

fork进程，在内核创建进程相关内核项，加载进程可执行文件；
查找依赖的.so，逐一加载映射虚拟地址；
初始化程序变量；

如下例通过strace查看pwd命令执行过程：

[root@~/wade/codeReview/learningc]# strace pwd
execve("/bin/pwd", ["pwd"], [/* 24 vars */]) = 0
brk(0)                                  = 0x1c77000
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb011000
access("/etc/ld.so.preload", R_OK)      = -1 ENOENT (No such file or directory)
open("/etc/ld.so.cache", O_RDONLY)      = 3
fstat(3, {st_mode=S_IFREG|0644, st_size=29690, ...}) = 0
mmap(NULL, 29690, PROT_READ, MAP_PRIVATE, 3, 0) = 0x7f06eb009000
close(3)                                = 0
open("/lib64/libc.so.6", O_RDONLY)      = 3
read(3, "\177ELF\2\1\1\3\0\0\0\0\0\0\0\0\3\0>\0\1\0\0\0p\356\1\0\0\0\0\0"..., 832) = 832
fstat(3, {st_mode=S_IFREG|0755, st_size=1921096, ...}) = 0
mmap(NULL, 3750152, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) = 0x7f06eaa5f000
mprotect(0x7f06eabe9000, 2097152, PROT_NONE) = 0
mmap(0x7f06eade9000, 20480, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x18a000) = 0x7f06eade9000
mmap(0x7f06eadee000, 18696, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0x7f06eadee000
close(3)                                = 0
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb008000
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb007000
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb006000
arch_prctl(ARCH_SET_FS, 0x7f06eb007700) = 0
mprotect(0x7f06eade9000, 16384, PROT_READ) = 0
mprotect(0x7f06eb012000, 4096, PROT_READ) = 0
munmap(0x7f06eb009000, 29690)           = 0
brk(0)                                  = 0x1c77000
brk(0x1c98000)                          = 0x1c98000
open("/usr/lib/locale/locale-archive", O_RDONLY) = 3
fstat(3, {st_mode=S_IFREG|0644, st_size=99158576, ...}) = 0
mmap(NULL, 99158576, PROT_READ, MAP_PRIVATE, 3, 0) = 0x7f06e4bce000
close(3)                                = 0
getcwd("/root/wade/codeReview/learningc", 4096) = 32
fstat(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 1), ...}) = 0
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb010000
write(1, "/root/wade/codeReview/learningc\n", 32/root/wade/codeReview/learningc
) = 32
close(1)                                = 0
munmap(0x7f06eb010000, 4096)            = 0
close(2)                                = 0
exit_group(0)                           = ?

由此可见，如果只加载必须的动态库对程序性能有很重要的意义。

最佳实践

1. 程序test.c：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
main () {
    printf ("1");
    exit (0);
}

2. 编译test.c

[root@~/wade/codeReview/learningc/16]# gcc -lm -lpthread -o test test.c

3. 通过ldd test命令查看程序运行时需要加载哪些动态库：

[root@~/wade/codeReview/learningc/16]# ldd test
	linux-vdso.so.1 =>  (0x00007fff6b5ff000)
	libm.so.6 => /lib64/libm.so.6 (0x00007f394cefa000)
	libpthread.so.0 => /lib64/libpthread.so.0 (0x00007f394ccdd000)
	libc.so.6 => /lib64/libc.so.6 (0x00007f394c948000)
	/lib64/ld-linux-x86-64.so.2 (0x00007f394d187000)

4. 通过ldd -u test命令查看程序加载了哪些不必要的动态库：

[root@~/wade/codeReview/learningc/16]# ldd -u test
Unused direct dependencies:
 
	/lib64/libm.so.6
	/lib64/libpthread.so.0

5. 通过-Wl,–as-needed编译选项只加载必须的动态库：

[root@~/wade/codeReview/learningc/16]# gcc -Wl,--as-needed -lm -o test test.c 
[root@~/wade/codeReview/learningc/16]# ldd test
	linux-vdso.so.1 =>  (0x00007fffdf5ff000)
	libc.so.6 => /lib64/libc.so.6 (0x00007f5d862cd000)
	/lib64/ld-linux-x86-64.so.2 (0x00007f5d8666a000)