在选择系统调用查询 [英] Query on Select System Call

问题描述

选择（）的定义是：

int select(int nfds, fd_set *readfds, fd_set *writefds, fd_set *errorfds, struct timeval *timeout);

的 NFDs的的再presents在所有已知集合加上一个最高的文件描述符。我想知道这是为什么数据需要选择（）时FD_SET资料。

nfds represents the highest file descriptor in all given sets plus one. I would like to know why is this data required for select() when the fd_set information is available.

如果该集合中的文件描述符是说，4，8，9，值的 NFDs的的将是10.将选择（）箴言报FDS 9,8,7,6,5,4 ？

If the FDs in the set are say, 4, 8, 9 ,the value of nfds would be 10. Would select() moniter fds 9,8,7,6,5,4 ?

推荐答案

美中不足的是，FD_SET是不是一个真正的设置的方式，你在想。在后面的幕后细节是一个的fd_set的实现只是用作位域的整数。换句话说，执行

The catch is that fd_set is not really a "set" in the way you're thinking. The behind-the-scenes detail is that the implementation of an fd_set is just an integer that is used as a bitfield. In other words, executing

fd_set foo;
FD_CLEAR(&foo);
FD_SET(&foo, 3);

设定为foo的十进制值8 - 它设置的第四最低singificant位为1（记住，0是一个有效的描述符）

Sets foo to decimal value 8 - it sets the fourth-least-singificant bit to 1 (remember that 0 is a valid descriptor).

FD_SET(&foo, 3);

等同于

foo |= (1 << 3);

因此​​，为了使选择工作的权利，它需要知道其中的fd_set的位是你关心位。否则就没有办法为它告诉一个零位即是中的集合，但是从零位是不在集合设置为false。

So in order for select to work right, it needs to know which bits of the fd_set are bits that you care about. Otherwise there would be no way for it to tell a zero bit that is "in" the set but set to false from a zero bit that is "not in" the set.

在您的例子中，有4个，8个和9套和n = 10的fd_set是PTED为一组10个条目（FDS 0-9）间$ P $。参赛作品4,8和9是真（监视它们）。参赛作品1,2,3,5,6,7-都是假的（不监视它们）。任何FD值大于9，根本就不是在设定的时间。

In your example, a fd_set with 4, 8, and 9 set and n = 10 is interpreted as "A set with 10 entries (fds 0-9). Entries 4, 8, and 9 are true (monitor them). Entries 1,2,3,5,6,7 are false (don't monitor them). Any fd value greater than 9 is simply not in the set period."

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