>/dev/null吃掉输出流吗? [英] How does > /dev/null eat up output streams?

问题描述

我在bash编程中经常使用/dev/null 将不必要的输出发送到黑洞中.

I've used /dev/null a lot in bash programming to send unnecessary output into a black hole.

例如，以下命令:

$ echo 'foo bar' > /dev/null
$ 

不会 echo 任何东西.我读过/dev/null 是一个空文件，用于通过重定向处理不需要的输出.但是，这种处置究竟是如何进行的呢?我无法想象/dev/null 将内容写入文件，然后立即删除该文件.那么，当您重定向到此文件时实际发生了什么?

Will not echo anything. I've read that /dev/null is an empty file used to dispose of unwanted output through redirection. But how exactly does this disposal take place? I can't imagine /dev/null writing the content to a file and then immediately deleting that file. So what actually happens when you redirect to this file?

推荐答案

>/dev/null 将命令标准输出重定向到null设备，这是一个特殊的设备，它将丢弃写入的信息做到这一点.所有这些都是通过file_operations实现的( drivers/char/mem.c 如果您想看一下自己的话):

>/dev/null redirects the command standard output to the null device, which is a special device which discards the information written to it. It's all implemented via file_operations (drivers/char/mem.c if you're curious to look yourself):

static const struct file_operations null_fops = {
    .llseek     = null_lseek,
    .read       = read_null,
    .write      = write_null,
    .splice_write   = splice_write_null,
};

write_null是您写入/dev/null时所调用的名称.它总是返回与您写入的字节数相同的字节数:

write_null is what's called when you write to /dev/null. It always returns the same number of bytes that you write to it:

static ssize_t write_null(struct file *file, const char __user *buf,
              size_t count, loff_t *ppos)
{
    return count;
}

就是这样.缓冲区将被忽略.

That's it. The buffer is just ignored.

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