If fclose(0) is called, does this close stdin?

问题:

If fclose(0) is called, does this close stdin?

The reason why I'm asking this is that for some reason, stdin is being closed in my application and I cannot figure out why. I checked for fclose (stdin) and this is not in the application and so I was wondering if fclose(0) could cause undefined behaviour such as closing stdin?

If not, what are other ways that stdin could be erroneously closed?

回答1:

The signature of fclose is this:

int fclose ( FILE * stream );

That means, fclose expects a pointer to FILE object. So if you pass 0, instead of a pointer, 0 would be understood as NULL pointer¹. If its NULL pointer, how do you expect it to close stdin? It will not close. Use fclose(stdin), as stdin itself is a pointer to FILE object.

I think you're confusing stdin with file-descriptor which is of integral type, and usually denoted as fd. Its true that fd of input stream is 0. So if you want to use fd (instead of FILE*), then you've to use close from <unistd.h>.

#include <unistd.h>
int close(int fildes);

That is, close(0) would close stdin.

_{1 : It seems interesting that if you had passed 1 to fclose with the intention to close stdout, your code wouldn't even compile, and you would immediately see the problem with your code at compile-time itself. Now the question is, why would it not compile? Because unlike 0, 1 is not implicitly converted into pointer type. The compiler would generate message like "error: invalid conversion from ‘int’ to ‘FILE*’. See the error and line number here at ideone.}

回答2:

I think it will take it as fclose(NULL); Which should be undefined and may crash.

回答3:

The following closes stdin: close(0); fclose(stdin); close(STDIN_FILENO); daemon(0, 0);

回答4:

fclose(0) invokes undefined behavior, so yes, it could do anything, including closing stdin. But you have much bigger problems if fclose(0) appears in your code.