c - fclose 似乎不能在 C 中工作

Question

    FILE *file = fopen(argv[1], "r");
    if(file != NULL){
        char command[MAX_BUFFER];
        while(fgets(command, MAX_BUFFER, file) != NULL){ //read line
            //operations
        }
        fclose(file);
    }
    else write(fileno(stderr), ERROR_MESSAGE, strlen(ERROR_MESSAGE));

我正在研究一个非常基本的 UNIX shell 实现，它从文件中逐行读取和执行。从上面的代码中，我试图找出为什么 fclose() 可能会失败，因为 valgrind 似乎告诉我我已经打开了这个文件描述符。

我假设泄漏摘要中“仍可访问”的 568 字节是指以某种方式失败的 fclose() 是否正确？

==25428== FILE DESCRIPTORS: 4 open at exit.
==25428== Open file descriptor 3: test
==25428==    at 0x4F186B0: __open_nocancel (syscall-template.S:82)
==25428==    by 0x4EAC628: _IO_file_fopen@@GLIBC_2.2.5 (fileops.c:233)
==25428==    by 0x4EA1265: __fopen_internal (iofopen.c:93)
==25428==    by 0x400C0C: main (in /home/Desktop/sh)
==25428== 
==25428== Open file descriptor 2: /dev/pts/0
==25428==    <inherited from parent>
==25428== 
==25428== Open file descriptor 1: /dev/pts/0
==25428==    <inherited from parent>
==25428== 
==25428== Open file descriptor 0: /dev/pts/0
==25428==    <inherited from parent>
==25428== 
==25428== LEAK SUMMARY:
==25428==    definitely lost: 0 bytes in 0 blocks
==25428==    indirectly lost: 0 bytes in 0 blocks
==25428==      possibly lost: 0 bytes in 0 blocks
==25428==    still reachable: 568 bytes in 1 blocks
==25428==         suppressed: 0 bytes in 0 blocks

Answer 1

对了，检查返回值，也贴一个perror来得到一个简短的解释。如果您想确认它仍然打开，请尝试 lsof，但您可能需要用 grep 获取程序名称或 pid，因为可能有几十个。