如何使用x86-64 Linux系统调用附加到文件? [英] How to append to a file using x86-64 Linux system calls?
问题描述
我正在创建一个文件，然后使用系统调用将其打开.
I'm creating a file and then opening it using system calls.
创建:
mov rax, 85
mov rdi, path
mov rsi, 400
syscall
开幕:
mov rax, 2
mov rdi, path
mov rsi, 2 ; I found somewhere that I should use 400 to append
syscall
mov r8, rax ; save the file handle
写作:
mov rax, 1
mov rdi, r8
mov rsi, output
mov rdx, length
syscall
完成这些步骤后，我自然关闭了文件，但是我无法追加文件，并且每次执行这些操作时，它只会重写文件的内容，但是例如，我计算了前n个素数并存储了将它们存储在一些内存中，现在我想在文件中写下这n个素数，但是如果不添加更多行就无法这样做.
After I complete these steps I close the file naturally, however I am unable to append to it and each time I perform these actions it just rewrites the content of the file, but for example I calculated the first n prime numbers and stored them away in some chunk of memory and now I'd like to write those n prime numbers down in the file, but I'm not able to do so without appending more lines.
还有一种方法可以不为rdx中的write指定字符串的长度，而是用/0终止字符串吗?
Also is there a way to not specify the length of the string for write in rdx, but instead have the string be terminated by a /0?
推荐答案
要附加到文件，请使用O_APPEND = 0x400标志作为rsi中的int flags参数.所以您接近了:它的值是十六进制的400，而不是您怀疑的十进制的400.
For appending to a file use the O_APPEND = 0x400 flag as int flags parameter in rsi. So you were close: it's a value of hexadecimal 400 and not decimal 400 as you suspected.
以下是十六进制原始int flags值的列表:
Here is a list of the raw int flags values in hexadecimal:
O_ACCMODE = 0x3
O_APPEND = 0x400
O_ASYNC = 0x2000
O_CLOEXEC = 0x80000
O_CREAT = 0x40
O_DIRECT = 0x4000
O_DIRECTORY = 0x10000
O_DSYNC = 0x1000
O_EXCL = 0x80
O_FSYNC = 0x101000
O_LARGEFILE = 0x0
O_NDELAY = 0x800
O_NOATIME = 0x40000
O_NOCTTY = 0x100
O_NOFOLLOW = 0x20000
O_NONBLOCK = 0x800
O_RDONLY = 0x0
O_RDWR = 0x2
O_RSYNC = 0x101000
O_SYNC = 0x101000
O_TRUNC = 0x200
O_WRONLY = 0x1
所以这应该起作用:
mov rax, 2
lea rdi, [rel path]
mov rsi, 0x441 ; O_CREAT| O_WRONLY | O_APPEND
mov edx, 0q666 ; octal permissions in case O_CREAT has to create it
syscall
mov r8, rax ; save the file descriptor
如果文件不存在，我添加了三个值来创建文件；如果文件存在，则以write_only模式打开它.您可以将它们定义为汇编时equ常量，以便实际上可以在源代码中编写mov esi, O_CREAT| O_WRONLY | O_APPEND.
I added the three values to create the file if it doesn't exist or open it in write_only mode if it does exist. You could define them as assemble-time equ constants so you could actually write mov esi, O_CREAT| O_WRONLY | O_APPEND in your source.
此外，请确保在每次包含O_CREAT时都为第三个arg(权限)提供一个值.否则，该文件可能会因其权限(可能包括setuid + exec)而被随机垃圾创建.通常，您传递八进制666(NASM 0q666)，然后让用户的umask取消对其他(可选)组的写入权限.
Also, make sure to supply a value for the 3rd arg (permissions) any time you include O_CREAT. Otherwise the file could get created with random garbage for its permissions, including possibly setuid + exec. Normally you pass octal 666 (NASM 0q666) and let the user's umask knock off write permission for other and optionally group.
如果您实际上不希望打开它(如果尚不存在)，请省略O_CREAT.然后它将返回-ENOENT.
If you don't actually want open to create it if it doesn't already exist, omit O_CREAT. Then it will return -ENOENT.
还有一种方法不指定rdx中字符串的长度，而是用/0终止字符串吗?
Also is there a way to not specify the length of the string in rdx, but instead have the string be terminated by a /0?
简而言之:No.
创建一个strlen函数，并将结果传递到rdx中.内核系统调用将要在文件描述符上读取/写入的数据作为缓冲区+长度，而仅将路径名作为C隐式长度字符串.
In short: No.
Create a strlen function and pass the result in rdx. Kernel system calls take data to be read/written on file descriptors as buffer + length, only pathnames as C implicit-length strings.
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