为什么 mmap 在 printfs 调用期间完成？

Question

为什么 printf() 执行 sys_mmap() 然后将字符串的内容以块（1024）复制到新地址 space对于 sys_write() ？

简单静态"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 的对象转储

> 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hook sys_write()系统调用的地址，我们看到传递给它的地址是mmap-ed 地址区域。考虑到字符串已经存在于二进制的第一个可加载段的 .rodata 部分中，这不仅仅是浪费新地址 space。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，可以通过更简单的方式进行验证。 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

您尝试过使用调试器吗？

$ 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]

所以它为缓冲输入输出分配内存，FILE* 使用。请注意，仅将 printf 与常量字符串一起使用将导致调用 puts，因为 GCC 足够智能。 puts(string) 实际上是一个 fputs(string, stdout)，其中标准输出是 FILE*。

使用原始写入，但不会导致此类行为：

#include <unistd.h>

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

为什么 mmap 在 printfs 调用期间完成？

Why is mmap done during printfs calls?

linux

linux-kernel

system

elf

libc