unix环境高级编程中一个习题的疑问

amazingjxq

题目：
The following sequence of code has been observed in various programs:
      dup2(fd, 0);
      dup2(fd, 1);
      dup2(fd, 2);
      if (fd > 2)
          close(fd);
To see why the if test is needed, assume that fd is 1 and draw a picture of what happens to the three descriptor entries and the corresponding file table entry with each call to dup2. Then assume that fd is 3 and draw the same picture.
给出的答案：
If fd is 1, then the dup2(fd, 1) returns 1 without closing descriptor 1. (Remember our discussion of this in Section 3.12.) After the three calls to dup2, all three descriptors point to the same file table entry. Nothing needs to be closed.
If fd is 3, however, after the three calls to dup2, four descriptors are pointing to the same file table entry. In this case, we need to close descriptor 3.

看答案的意思难道指向同一个文件表项的文件描述符个数有限制么？

nelab

有限制，但肯定不是3，
这里的意思应该是只保留012，其他的关闭

icebox

以前看到过有些程序里面有这段代码，目的是要将012这三个标准设备重定向到fd，完了之后，fd就没有打开的必要，但是在关闭的时候必须检查fd的值，不能close掉012这三个。

bobozhang

问题的关键不在于非得关闭那个fd，而在于如果要关闭，需要判断 fd > 2