Question

我正在阅读glibc-2.19的来源。我发现如果我使用fopen来获取FILE的指针，则stdio的缓冲区已经存在。它何时何地分配和初始化？

Answer 1

我发现如果我使用fopen获取FILE指针，stdio的缓冲区已经存在。

目前尚不清楚您的确切内容或方式。通常情况下，FILE使用的缓冲区不分配，直到您尝试读取或写入FILE的内容。

示例：

#include <stdio.h>

int main()
{
  FILE *fp = fopen("/etc/passwd", "r");
  int c = fgetc(fp);
  return 0;
}

gcc -g t.c && gdb -q ./a.out
Reading symbols from ./a.out...done.
(gdb) start
Temporary breakpoint 1 at 0x400535: file t.c, line 5.
Starting program: /tmp/a.out

Temporary breakpoint 1, main () at t.c:5
5     FILE *fp = fopen("/etc/passwd", "r");
(gdb) n
6     int c = fgetc(fp);
(gdb) p *fp
$1 = {
  _flags = -72539000,
  _IO_read_ptr = 0x0,
  _IO_read_end = 0x0,
  _IO_read_base = 0x0,
  _IO_write_base = 0x0,
  _IO_write_ptr = 0x0,
  _IO_write_end = 0x0,
  _IO_buf_base = 0x0,
  _IO_buf_end = 0x0,
  _IO_save_base = 0x0,
  _IO_backup_base = 0x0,
  _IO_save_end = 0x0,
  _markers = 0x0,
  _chain = 0x7ffff7dd41c0 <_IO_2_1_stderr_>,
  _fileno = 3,
  _flags2 = 0,
  _old_offset = 0,
  _cur_column = 0,
  _vtable_offset = 0 '\000',
  _shortbuf = "",
  _lock = 0x6020f0,
  _offset = -1,
  __pad1 = 0x0,
  __pad2 = 0x602100,
  __pad3 = 0x0,
  __pad4 = 0x0,
  __pad5 = 0,
  _mode = 0,
  _unused2 = '\000' <repeats 19 times>
}

上面你可以清楚地看到，还没有分配任何内部缓冲区：_IO_read_ptr，_IO_read_end等。

现在让我们在&fp->_IO_read_ptr和next上设置一个观察点：

(gdb) watch -l fp._IO_read_ptr
Hardware watchpoint 3: -location fp._IO_read_ptr
(gdb) next
Hardware watchpoint 3: -location fp._IO_read_ptr

Old value = 0x0
New value = 0x7ffff7ff7000 ""
0x00007ffff7a8f689 in _IO_new_file_underflow (fp=0x602010) at fileops.c:608
608 fileops.c: No such file or directory.
(gdb) bt
#0  0x00007ffff7a8f689 in _IO_new_file_underflow (fp=0x602010) at fileops.c:608
#1  0x00007ffff7a9062e in __GI__IO_default_uflow (fp=0x602010) at genops.c:435
#2  0x00007ffff7a86bae in _IO_getc (fp=0x602010) at getc.c:39
#3  0x00000000004005a4 in main () at t.c:6

现在您可以看到尝试从FILE读取确实会导致缓冲区在_IO_new_file_underflow中分配。

glibc：分配和初始化stdio流缓冲区的时间和位置是什么？

1 个答案: