开发者

fopen without fclose in C

开发者 https://www.devze.com 2023-01-09 00:22 出处:网络
What happens if i open a file using fopen some n number of times without calling fclose on it? Any buffer overrun issues m开发者_运维技巧ay arise?If you continue to fopen without fclose then eventual

What happens if i open a file using fopen some n number of times without calling fclose on it?

Any buffer overrun issues m开发者_运维技巧ay arise?


If you continue to fopen without fclose then eventually your future calls to fopen will start to fail. There are a limited number of file descriptors available to your program.

See this related SO question.


You waste underlying file handles. Please close any files you open, in a timely fashion, as to avoid this leak of resources.


If you continue opening files without closing them then you will run out of file descriptors at some point, either at the application level or the OS level, and all further attempts to open a file will fail.


Aside from wasting file descriptors for the process as others answered, you would also waste memory since each file stream manages in/out buffers that are allocated internally by libc.


As others have mentioned, you don't want to leak file descriptors. But it is ok to have a single file open multiple times. The file descriptors are independent and will not interfere with each other (assuming you are just reading and not writing to the file).

#include <stdio.h>

int main()
{
    FILE* f1 = fopen("/tmp/foo.txt", "rb");
    FILE* f2 = fopen("/tmp/foo.txt", "rb");
    FILE* f3 = fopen("/tmp/foo.txt", "rb");
    FILE* f4 = fopen("/tmp/foo.txt", "rb");

    char buf1[32] = { 0, };
    char buf2[32] = { 0, };
    char buf3[32] = { 0, };
    char buf4[32] = { 0, };

    fread(buf1, 1, sizeof(buf1) - 1, f1);
    fread(buf2, 1, sizeof(buf2) - 1, f2);
    fread(buf3, 1, sizeof(buf3) - 1, f3);
    fread(buf4, 1, sizeof(buf4) - 1, f4);

    printf("buf1 = '%s'\n", buf1);
    printf("buf2 = '%s'\n", buf2);
    printf("buf3 = '%s'\n", buf3);
    printf("buf4 = '%s'\n", buf4);

    fclose(f1);
    fclose(f2);
    fclose(f3);
    fclose(f4);

    return 0;
}

Gives output like:

$ ./fopen 
buf1 = '0123456789ABCDEFGHIJ0123456789a'
buf2 = '0123456789ABCDEFGHIJ0123456789a'
buf3 = '0123456789ABCDEFGHIJ0123456789a'
buf4 = '0123456789ABCDEFGHIJ0123456789a'
0

精彩评论

暂无评论...
验证码 换一张
取 消