【发布时间】:2019-04-05 02:09:16
【问题描述】:
man writev 是这样说的:
readv() 和 writev() 执行的数据传输是原子的:writev() 写入的数据作为单个块写入,不会与其他进程的写入输出混合(但请参见 pipe(7)例外);类似地,readv() 是有保证的
这是来自man 7 pipe:
O_NONBLOCK disabled, n <= PIPE_BUF All n bytes are written atomically; write(2) may block if there is not room for n bytes to be written immediately O_NONBLOCK enabled, n <= PIPE_BUF If there is room to write n bytes to the pipe, then write(2) succeeds immediately, writing all n bytes; otherwise write(2) fails, with errno set to EAGAIN. O_NONBLOCK disabled, n > PIPE_BUF The write is nonatomic: the data given to write(2) may be interleaved with write(2)s by other process; the write(2) blocks until n bytes have been written. O_NONBLOCK enabled, n > PIPE_BUF If the pipe is full, then write(2) fails, with errno set to EAGAIN. Otherwise, from 1 to n bytes may be written (i.e., a "partial write" may occur; the caller should check the return value from write(2) to see how many bytes were actually written), and these bytes may be interleaved with writes by other processes.
$ cat writev.c
#include <string.h>
#include <sys/uio.h>
int
main(int argc,char **argv) {
static char part1[] = "ST";
static char part2[] = "\n";
struct iovec iov[2];
iov[0].iov_base = part1;
iov[0].iov_len = strlen(part1);
iov[1].iov_base = part2;
iov[1].iov_len = strlen(part2);
writev(1,iov,2);
return 0;
}
$ gcc writev.c
$ unbuffer bash -c 'for ((i=0; i<50; i++)); do ./a.out & ./a.out; done' | wc -c
300 # < PIPE_BUF
# Run the following several times to get the output corrupted
$ unbuffer bash -c 'for ((i=0; i<50; i++)); do ./a.out & ./a.out; done' | sort | uniq -c
4
92 ST
4 STST
如果 writev 是原子的(根据文档),谁能解释为什么不同写入的输出是交错的?
更新:
来自strace -fo /tmp/log unbuffer bash -c 'for ((i=0; i<10000; i++)); do ./a.out & ./a.out; done' | sort | uniq -c的一些相关数据
13301 writev(1, [{iov_base="ST", iov_len=2}, {iov_base="\n", iov_len=1}], 2 <unfinished ...>
13302 mprotect(0x56397d7d8000, 4096, PROT_READ) = 0
13302 mprotect(0x7f7190c68000, 4096, PROT_READ) = 0
13302 munmap(0x7f7190c51000, 90695) = 0
13302 writev(1, [{iov_base="ST", iov_len=2}, {iov_base="\n", iov_len=1}], 2) = 3
13301 <... writev resumed> ) = 3
24814 <... select resumed> ) = 1 (in [4])
13302 exit_group(0 <unfinished ...>
13301 exit_group(0 <unfinished ...>
13302 <... exit_group resumed>) = ?
13301 <... exit_group resumed>) = ?
24814 futex(0x55b5b8c11cc4, FUTEX_WAKE_PRIVATE, 2147483647 <unfinished ...>
24807 <... futex resumed> ) = 0
24814 <... futex resumed> ) = 1
24807 futex(0x7f7f55e8f920, FUTEX_WAIT_PRIVATE, 2, NULL <unfinished ...>
13302 +++ exited with 0 +++
24807 <... futex resumed> ) = -1 EAGAIN (Resource temporarily unavailable)
13301 +++ exited with 0 +++
24807 futex(0x7f7f55e8f920, FUTEX_WAKE_PRIVATE, 1 <unfinished ...>
24814 futex(0x7f7f55e8f920, FUTEX_WAKE_PRIVATE, 1 <unfinished ...>
24807 <... futex resumed> ) = 0
24814 <... futex resumed> ) = 0
24807 read(4, <unfinished ...>
24814 select(6, [5], [], [], NULL <unfinished ...>
24807 <... read resumed> "STST\n\n", 4096) = 6
24808 <... wait4 resumed> [{WIFEXITED(s) && WEXITSTATUS(s) == 0}], 0, NULL) = 13302
24807 write(1, "STST\n\n", 6 <unfinished ...>
【问题讨论】:
-
如果您不使用
stdio,则无需使用unbuffer。 -
我在没有
unbuffer的情况下尝试过:for ((i=0; i<50; i++)); do ./a.out & ./a.out; done | sort | uniq -c并在 Debian 3.2.89-2 上获得了100 ST。 -
这肯定是由
unbuffer正在做的事情引起的。 -
您要求它在命令中写入 2 个缓冲区。我认为,单个缓冲区的写入是原子的。但是不同的缓冲区可以交错。你的输出支持这个理论。在他的文档中也暗示了这一点。
-
@user207421:程序并没有真正写入管道。使用
unbuffer,程序a.out的fd 1 进入伪tty,并且有一个辅助进程收集写入pty 的数据并将其写入管道。这个想法是,这将鼓励使用 stdio 到行缓冲区的程序。因此,终端似乎是该规则的另一个(未记录的)例外。
标签: c++ c linux-kernel system-calls