有人可以解释一下stdio缓冲的工作原理吗? [英] Can someone please explain how stdio buffering works?
问题描述
我不知道缓冲区在做什么以及如何使用.(此外,如果您可以解释缓冲区的正常功能)特别是为什么在此示例中需要使用fflush?
I don't understand what the buffer is doing and how it's used. (Also, if you can explain what a buffer normally does) In particular, why do I need fflush in this example?
int main(int argc, char **argv)
{
int pid, status;
int newfd; /* new file descriptor */
if (argc != 2) {
fprintf(stderr, "usage: %s output_file\n", argv[0]);
exit(1);
}
if ((newfd = open(argv[1], O_CREAT|O_TRUNC|O_WRONLY, 0644)) < 0) {
perror(argv[1]); /* open failed */
exit(1);
}
printf("This goes to the standard output.\n");
printf("Now the standard output will go to \"%s\".\n", argv[1]);
fflush(stdout);
/* this new file will become the standard output */
/* standard output is file descriptor 1, so we use dup2 to */
/* to copy the new file descriptor onto file descriptor 1 */
/* dup2 will close the current standard output */
dup2(newfd, 1);
printf("This goes to the standard output too.\n");
exit(0);
}
推荐答案
在UNIX系统中,stdout缓冲可以提高I/O性能.每次执行I/O都将非常昂贵.
In a UNIX system the stdout buffering happens to improve I/O performance. It would be very expensive to do I/O every time.
如果您真的不想缓冲,则有一些选择:
If you really don't want to buffer there's some options:
-
禁用缓冲调用 setvbuf http://www.cplusplus.com/reference/cstdio/setvbuf/
要刷新缓冲区时调用flush
Call flush when you want to flush the buffer
输出到stderr(默认情况下为无缓冲)
Output to stderr (that's unbuffered by default)
这里有更多详细信息: http://www.turnkeylinux.org/blog/unix-缓冲
Here you've more details: http://www.turnkeylinux.org/blog/unix-buffering
I/O是一项昂贵的操作,因此,为了减少I/O操作的数量,系统会将信息存储在临时存储位置中,并将I/O操作延迟到其具有大量数据的时刻
I/O is an expensive operation, so to reduce the number of I/O operations the system store the information in a temporary memory location, and delay the I/O operation to a moment when it has a good amount of data.
这样,您的I/O操作数量少得多,这意味着应用程序速度更快.
This way you've a much smaller number of I/O operations, what means, a faster application.
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