在运行 Ubuntu 19.10 的 x64 系统中，文件描述符的大小是多少？

Question

文件描述符（例如标准输入和标准输出）的位大小是多少，它是 32 位整数吗？

Answer 1

If you are talking about the actual file descriptors returned by (and used for) Linux syscalls, then take a look at the manpage for open etc. as @JonathanLeffler suggests.

For instance:

 int open(const char *pathname, int flags);

The return value of open() is a file descriptor, a small, nonnegative integer that is used in subsequent system calls. [...] The file descriptor returned by a successful call will be the lowest-numbered file descriptor not currently open for the process.

Given that Unix-like systems are LP64, int and therefore the file descriptors are 32-bit wide.

However, note that the kernel will give you as small integers as possible and that you will typically reach a limit way before that (see Limits on the number of file descriptors ) either due to the kernel global limit or the soft/hard limits.

This means that, if you really needed it, you could in theory use a smaller integer to store your file descriptors, eg an int16_t or an int8_t (assuming that your process does not use that many file descriptors at a time).

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If, instead, you are referring to stdin etc., those are not file descriptors but file streams defined by the C standard.

They are macros which expand to expressions with pointer type ( FILE * ), and pointers in a typical 64-bit platform like x86_64 are 64-bit wide.

See 7.21p3 (Input/output <stdio.h> ):

 stdin stdout stderr

which are expressions of type ''pointer to FILE '' that point to the FILE objects associated, respectively, with the standard error, input, and output streams.