我在多个地方读到,管道的默认缓冲区大小为4kB(例如,在此处),我ulimit -a倾向于确认该语句:
ulimit -a
$ ulimit -a core file size (blocks, -c) 0 data seg size (kbytes, -d) unlimited scheduling priority (-e) 0 file size (blocks, -f) unlimited pending signals (-i) 15923 max locked memory (kbytes, -l) 64 max memory size (kbytes, -m) unlimited open files (-n) 1024 pipe size (512 bytes, -p) 8 // 8 * 512B = 4kB POSIX message queues (bytes, -q) 819200 real-time priority (-r) 0 stack size (kbytes, -s) 10240 cpu time (seconds, -t) unlimited max user processes (-u) 1024 virtual memory (kbytes, -v) unlimited file locks (-x) unlimited
但是,当我使用一个小程序来测试缓冲区大小时(通过写入shell直到write()块),我看到了64kB的限制!
看到这个程序:
#include <stdio.h> #include <unistd.h> #include <limits.h> int main(void) { int tube[2]; char c = 'c'; int i; fprintf(stdout, "Tube Creation\n"); fprintf(stdout, "Theoretical max size: %d\n", PIPE_BUF); if( pipe(tube) != 0) { perror("pipe"); _exit(1); } fprintf(stdout, "Writing in pipe\n"); for(i=0;; i++) { fprintf(stdout, "%d bytes written\n", i+1); if( write(tube[1], &c, 1) != 1) { perror("Write"); _exit(1); } } return 0; }
及其输出:
$ ./test_buf_pipe Tube Creation Theoretical max size: 4096 Writing in pipe 1 bytes written 2 bytes written 3 bytes written 4 bytes written [...] 65535 bytes written [blocks here]
强烈建议shell缓冲区的大小实际上是64k!这是怎么回事?
其他答案告诉您管道大小为64 KB。PIPE_BUF为4KB的原因是PIPE_BUF是最大的大小,可以保证对其进行原子写操作。参见http://pubs.opengroup.org/onlinepubs/9699919799/functions/write.html