前几日，我再看书的时候，仔细的回顾了关于信号的这一章，当时，我想到了如果是在调用系统调用，这个时候给进程发发送一个信号，会产生什么效果，很多人都说了，就算是系统调用，他会很快的执行完成的，也是，Linux信号有一种机制就是延时投递，比如这时，你的进程正在malloc分配内存，或者执行其他的原子操作，但是会很快的执行完成，这就会触发这种机制！
但是，如果我们是阻塞方式的socket套接字调用read，或者recv，recvfrom等，这会一直阻塞，难道非得等到有数据来到，我们才能给他信号？显然Linux内核人员避免了这种情况，还有，如果程序就是在sleep，成为了睡眠状态，这时候给他一个信号，会产生什么效果？
我是根据一个程序是在sleep的时候，接受到了一个信号想到了许多东西，下面看一下这个程序：
#include <stdio.h>
#include <signal.h>
void
handler (int signum)
{
printf ("signal called\n");
}
int
main ()
{
struct sigaction sigact;
sigact.sa_handler = handler;
//sigact.sa_flags = SA_RESTART;
sigaction (SIGALRM, &sigact, NULL);
alarm (2);
sleep (50);
printf ("done\n");
return 0;
}
这个程序是很简单的，当你运行的时候就知道了结果是什么了，当然，我并没有满足这点内容，下面，我又用了strace -o log.txt ./a.out运行来查看这个程序运行了什么系统调用（内核层面），当你执行完后，目录下会多了一个log.txt文件，打开它，你会看到如下：
1 execve("./a.out", ["./a.out"], [/* 41 vars */]) = 0
2 brk(0) = 0x9ddd000
3 access("/etc/ld.so.nohwcap", F_OK) = -1 ENOENT (No such file or directory)
4 mmap2(NULL, 8192, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb77ea000
5 access("/etc/ld.so.preload", R_OK) = -1 ENOENT (No such file or directory)
6 open("/etc/ld.so.cache", O_RDONLY) = 3
7 fstat64(3, {st_mode=S_IFREG|0644, st_size=55715, ...}) = 0
8 mmap2(NULL, 55715, PROT_READ, MAP_PRIVATE, 3, 0) = 0xb77dc000
9 close(3) = 0
10 access("/etc/ld.so.nohwcap", F_OK) = -1 ENOENT (No such file or directory)
11 open("/lib/libc.so.6", O_RDONLY) = 3
12 read(3, "\177ELF\1\1\1\0\0\0\0\0\0\0\0\0\3\0\3\0\1\0\0\0@n\1\0004\0\0\0"..., 512) = 512
13 fstat64(3, {st_mode=S_IFREG|0755, st_size=1421892, ...}) = 0
14 mmap2(NULL, 1427880, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) = 0x7a6000
15 mmap2(0x8fd000, 12288, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x157) = 0x8fd000
16 mmap2(0x900000, 10664, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0x900000
17 close(3) = 0
18 mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb77db000
19 set_thread_area({entry_number:-1 -> 6, base_addr:0xb77db6c0, limit:1048575, seg_32bit:1, contents:0, read_exec_only:0, limit_in_pages:1, seg_not_present:0, useable:1}) = 0
20 mprotect(0x8fd000, 8192, PROT_READ) = 0
21 mprotect(0x8049000, 4096, PROT_READ) = 0
22 mprotect(0x24e000, 4096, PROT_READ) = 0
23 munmap(0xb77dc000, 55715) = 0
24 rt_sigaction(SIGALRM, {0x8048454, ~[HUP QUIT ILL TRAP ABRT BUS SEGV TERM STOP TSTP TTIN URG XCPU XFSZ VTALRM PROF WINCH IO PWR SYS RTMIN RT_1 RT_2 RT_4 RT_5 RT_10 RT_11 RT_12 RT_13 RT_14 RT_18 RT_19 RT_22 RT_23 RT_31], SA_STACK|SA_NOCLDSTOP|SA_NOCLDWAIT|0x484e8}, NULL, 8) = 0
25 alarm(2) = 0
26 rt_sigprocmask(SIG_BLOCK, [CHLD], [], 8) = 0
27 rt_sigaction(SIGCHLD, NULL, {SIG_DFL, [], 0}, 8) = 0
28 rt_sigprocmask(SIG_SETMASK, [], NULL, 8) = 0
29 nanosleep({50, 0}, {48, 195199}) = ? ERESTART_RESTARTBLOCK (To be restarted)
30 --- SIGALRM (Alarm clock) @ 0 (0) ---
31 fstat64(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 0), ...}) = 0
32 mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb77e9000
33 write(1, "signal called\n", 14) = 14
34 sigreturn() = ? (mask now [])
35 write(1, "done\n", 5) = 5
36 exit_group(0) = ?
对于我们来说，这个程序运行所调用的函数，会比你想的要多，其中，有很多的mmap2函数执行，我们再看第11行，这个是关于加载动态库的，也就是在关闭这个3文件描述符，都是在加载一些动态库，这些mmap2加载的地址勾起了我的兴趣，我们知道程序有什么堆栈，什么代码段，文本段，关于有什么段，《深入理解计算机系统》有详解，于是，我又修改了一下这个程序：
#include <stdio.h>
#include <signal.h>
#include <stdlib.h>
int gi1 = 1;
int gi2;
void
handler (int signum)
{
printf ("signal called\n");
printf ("handler function address is [%p]\n", handler);
return;
}
int
main ()
{
int i1 = 1;
int i2;
int *mi = (int *) malloc (10);
struct sigaction sigact;
sigact.sa_handler = handler;
//sigact.sa_flags = SA_RESTART;
sigaction (SIGALRM, &sigact, NULL);
alarm (2);
sleep (50);
printf ("done\n");
printf ("main function is [%p]\n", main);
printf ("gi1 is [%p]\n", &gi1);
printf ("gi2 is [%p]\n", &gi2);
printf ("i1 is [%p]\n", &i1);
printf ("i2 is [%p]\n", &i2);
printf ("mi is [%p]\n", mi);
free (mi);
return 0;
}
接着执行strace -o log2.txt ./a.out, 然后在打开看看这个文件：
1 execve("./a.out", ["./a.out"], [/* 41 vars */]) = 0
2 brk(0) = 0x9bae000
3 access("/etc/ld.so.nohwcap", F_OK) = -1 ENOENT (No such file or directory)
4 mmap2(NULL, 8192, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb7734000
5 access("/etc/ld.so.preload", R_OK) = -1 ENOENT (No such file or directory)
6 open("/etc/ld.so.cache", O_RDONLY) = 3
7 fstat64(3, {st_mode=S_IFREG|0644, st_size=55715, ...}) = 0
8 mmap2(NULL, 55715, PROT_READ, MAP_PRIVATE, 3, 0) = 0xb7726000
9 close(3) = 0
10 access("/etc/ld.so.nohwcap", F_OK) = -1 ENOENT (No such file or directory)
11 open("/lib/libc.so.6", O_RDONLY) = 3
12 read(3, "\177ELF\1\1\1\0\0\0\0\0\0\0\0\0\3\0\3\0\1\0\0\0@n\1\0004\0\0\0"..., 512) = 512
13 fstat64(3, {st_mode=S_IFREG|0755, st_size=1421892, ...}) = 0
14 mmap2(NULL, 1427880, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) = 0x426000
15 mmap2(0x57d000, 12288, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x157) = 0x57d000
16 mmap2(0x580000, 10664, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0x580000
17 close(3) = 0
18 mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb7725000
19 set_thread_area({entry_number:-1 -> 6, base_addr:0xb77256c0, limit:1048575, seg_32bit:1, contents:0, read_exec_only:0, limit_in_pages:1, seg_not_present:0, useable:1}) = 0
20 mprotect(0x57d000, 8192, PROT_READ) = 0
21 mprotect(0x8049000, 4096, PROT_READ) = 0
22 mprotect(0xfeb000, 4096, PROT_READ) = 0
23 munmap(0xb7726000, 55715) = 0
24 brk(0) = 0x9bae000
25 brk(0x9bcf000) = 0x9bcf000
26 rt_sigaction(SIGALRM, {0x8048620, [QUIT ILL TRAP BUS FPE KILL STKFLT STOP WINCH RT_1], SA_SIGINFO|0x57f320}, NULL, 8) = 0
27 alarm(2) = 0
28 rt_sigprocmask(SIG_BLOCK, [CHLD], [], 8) = 0
29 rt_sigaction(SIGCHLD, NULL, {SIG_DFL, [], 0}, 8) = 0
30 rt_sigprocmask(SIG_SETMASK, [], NULL, 8) = 0
31 nanosleep({50, 0}, {48, 188320}) = ? ERESTART_RESTARTBLOCK (To be restarted)
32 --- SIGALRM (Alarm clock) @ 0 (0) ---
33 fstat64(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 0), ...}) = 0
34 mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb7733000
35 write(1, "signal called\n", 14) = 14
36 write(1, "handler function address is [0x8"..., 40) = 40
37 rt_sigreturn(0xbfd456d4) = -1 EINTR (Interrupted system call)
38 write(1, "done\n", 5) = 5
39 write(1, "main function is [0x80484e0]\n", 29) = 29
40 write(1, "gi1 is [0x804a028]\n", 19) = 19
41 write(1, "gi2 is [0x804a034]\n", 19) = 19
42 write(1, "i1 is [0xbfd4579c]\n", 19) = 19
43 write(1, "i2 is [0xbfd45798]\n", 19) = 19
44 write(1, "mi is [0x9bae008]\n", 18) = 18
45 exit_group(0) = ?
前面的还是加载动态库，我们不用关心，当然，在第25行，程序执行了是brk函数，这个就是库向内核申请内存，但是内核给库函数最少就是一页（4096字节），如果用户申请少于这个，那就是有C 库分配给用户进程了！而在第35和36是程序信号输出，我们看到了一个地址开头就是8，我们就可已放心的认定，Linux的信号是执行在用户地址空间的栈上的，我们都知道，在IA-32体系下的计算机，linux内核虚拟地址空间是从0到0xC0000000，每个用户都有3G可用，而内存映射的地址是起始于0x40000000，这就说堆只有一个G可以用，继续增长就会到了mmap区域，然后Linux内核2.6.7版本开发的时候是将mmap地址上移动，这种解决方式就是《计算机程序设计艺术》中第一卷，第二章的顺序分配中的一个解决方法，程序划分各个段的目的，也就是说A
程序和B 程序的text段可能是挨着的，也就是说，你的程序不是在一块连续的内存上的，这也符合了程序的局部性原理！
好了，关于另外两种情况大家自己可以试试！