LinuxDLL加载优化方案-白红宇

LinuxDLL加载优化方案

阅读量：6670 次

发布时间：2019-06-25

本文共 3722 字，大约阅读时间需要 12 分钟。

作者：zhanhailiang 日期：2014-10-26

linux程序动态库载入流程简单介绍

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

fork进程，在内核创建进程相关内核项，载入进程可运行文件；

查找依赖的.so，逐一载入映射虚拟地址；

初始化程序变量；

例如以下例通过strace查看pwd命令运行过程：

[root@~/wade/codeReview/learningc]# strace pwdexecve("/bin/pwd", ["pwd"], [/* 24 vars */]) = 0brk(0)                                  = 0x1c77000mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb011000access("/etc/ld.so.preload", R_OK)      = -1 ENOENT (No such file or directory)open("/etc/ld.so.cache", O_RDONLY)      = 3fstat(3, {
     st_mode=S_IFREG|0644, st_size=29690, ...}) = 0mmap(NULL, 29690, PROT_READ, MAP_PRIVATE, 3, 0) = 0x7f06eb009000close(3)                                = 0open("/lib64/libc.so.6", O_RDONLY)      = 3read(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) = 832fstat(3, {
     st_mode=S_IFREG|0755, st_size=1921096, ...}) = 0mmap(NULL, 3750152, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) = 0x7f06eaa5f000mprotect(0x7f06eabe9000, 2097152, PROT_NONE) = 0mmap(0x7f06eade9000, 20480, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x18a000) = 0x7f06eade9000mmap(0x7f06eadee000, 18696, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0x7f06eadee000close(3)                                = 0mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb008000mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb007000mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb006000arch_prctl(ARCH_SET_FS, 0x7f06eb007700) = 0mprotect(0x7f06eade9000, 16384, PROT_READ) = 0mprotect(0x7f06eb012000, 4096, PROT_READ) = 0munmap(0x7f06eb009000, 29690)           = 0brk(0)                                  = 0x1c77000brk(0x1c98000)                          = 0x1c98000open("/usr/lib/locale/locale-archive", O_RDONLY) = 3fstat(3, {
     st_mode=S_IFREG|0644, st_size=99158576, ...}) = 0mmap(NULL, 99158576, PROT_READ, MAP_PRIVATE, 3, 0) = 0x7f06e4bce000close(3)                                = 0getcwd("/root/wade/codeReview/learningc", 4096) = 32fstat(1, {
     st_mode=S_IFCHR|0620, st_rdev=makedev(136, 1), ...}) = 0mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f06eb010000write(1, "/root/wade/codeReview/learningc\n", 32/root/wade/codeReview/learningc) = 32close(1)                                = 0munmap(0x7f06eb010000, 4096)            = 0close(2)                                = 0exit_group(0)                           = ?

由此可见。假设仅仅载入必须的动态库对程序性能有非常重要的意义。

最佳实践

1. 程序test.c：

#include 
      
       #include 
       
        #include 
        
          main () {
          ("1");     (0);}

2. 编译test.c

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

3. 通过ldd test命令查看程序执行时须要载入哪些动态库：

[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命令查看程序载入了哪些不必要的动态库：

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

5. 通过-Wl,–as-needed编译选项仅仅载入必须的动态库：

[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)

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