为什么在 printfs 调用期间完成 mmap？

Question

为什么 printf() 做一个 sys_mmap() 然后将字符串的内容以块（1024）的形式复制到 sys_write() 的新地址空间？

简单的静态“hello”程序的Strace如下所示。

> gcc -o hello -static hello.c
> strace ./hello


execve("./hello", ["./hello"], [/* 71 vars */]) = 0
uname({sys="Linux", node="Kumar", ...})   = 0
brk(0)                                  = 0x1ce8000
brk(0x1ce91c0)                          = 0x1ce91c0
arch_prctl(ARCH_SET_FS, 0x1ce8880)      = 0
readlink("/proc/self/exe", "/home/admin/hello", 4096) = 18
brk(0x1d0a1c0)                          = 0x1d0a1c0
brk(0x1d0b000)                          = 0x1d0b000
access("/etc/ld.so.nohwcap", F_OK)      = -1 ENOENT (No such file or directory)
fstat(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 28), ...}) = 0
mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7feda2130000
write(1, "Hello", 5Hello)                    = 5
exit_group(0)                           = ?
+++ exited with 0 +++

rodata的Objdump

> objdump -s --start-address=0x4935a0 ./hello | head -5

./hello:     file format elf64-x86-64

Contents of section .rodata:
 4935a0 01000200 48656c6c 6f006c69 62632d73  ....Hello.libc-s

如果我们在内核级别挂钩 sys_write() 系统调用的地址，我们会看到传递给它的地址是 mmap 的地址区域。考虑到字符串已经在二进制文件的第一个可加载段的 .rodata 部分中存在，这不仅仅是浪费新的地址空间。它与没有写权限等有关吗？那么为什么不让编译器首先将字符串放在 .data 部分（也是可写的）？

更新：

Mmap-ed 地址确实适用于 sys_write()，当我们将字符串变大（比如 ~1500 个字符）时，可以更轻松地验证它。 GDB会确认正在打印的数据地址【注意第二个断点】

(gdb) c
Continuing.
Hello World hhhhhhhhhhalhfafeuirafheuhrgiegieguehguergjkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqwwwwwwwwwwwwwwwwwwwwww     pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuiiiiiiiiiiiiiiiiiiiiiiiiiiiiiwqiuwqiuwiquwiqhchasnvjnavjanvjdanvjdanvjdanjfanvjaddijuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuquweuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuunnnnnnnnnnnnnnnnnnnnnnnnnnnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz,,,,,,,,,,,,,,,,,,,,,,
Breakpoint 1, _IO_new_file_write (f=0x6b8300 <_IO_2_1_stdout_>, data=0x7ffff7ffc000, n=706) at fileops.c:1257
1257    {

Answer 1

您是否尝试过使用调试器？

$ gdb /tmp/hello
...
(gdb) b __mmap
Breakpoint 1 at 0x4152e0    
(gdb) r
Starting program: /tmp/hello 

Breakpoint 1, 0x00000000004152e0 in mmap64 ()
(gdb) bt
#0  0x00000000004152e0 in mmap64 ()
#1  0x000000000045d73c in _IO_file_doallocate ()
#2  0x0000000000401fec in _IO_doallocbuf ()
#3  0x000000000042ca10 in _IO_new_file_overflow ()
#4  0x000000000042be9d in _IO_new_file_xsputn ()
#5  0x000000000040111d in puts ()
#6  0x00000000004004de in main () at hello.c:4
(gdb) c
Continuing.
Hello, w
[Inferior 1 (process 4294) exited with code 011]

因此它为FILE* 使用的缓冲输入输出分配内存。请注意，仅使用常量字符串的 printf 将导致 puts 调用，因为 GCC 足够聪明。而puts(string) 实际上是一个fputs(string, stdout)，其中stdout 是FILE*。

使用原始写入，但是不会产生这样的行为：

#include <unistd.h>

int main() {
    write(1, "Hello, w\n", sizeof("Hello, w\n"));
}