为什么在gdb中调用calloc不会将内存清零？

Question

我正在做一些编辑进程内存的实验，当它运行时，我注意到当我在gdb'd进程中调用calloc时，调用似乎工作并返回原始传递的指针，但内存似乎不被初始化为0：如果我叫memset的结果指针

(gdb) call calloc(1, 32) 
$88 = (void *) 0x8d9d50 
(gdb) x/8xw 0x8d9d50 
0x8d9d50:  0xf74a87d8  0x00007fff  0xf74a87d8  0x00007fff 
0x8d9d60:  0xfbfbfbfb  0xfbfbfbfb  0x00000000  0x9b510000 

，但是，初始化工作得很好：

(gdb) call memset(0x8d9d50, 0, 32) 
$89 = 9280848 
(gdb) x/8xw 0x8d9d50 
0x8d9d50:  0x00000000  0x00000000  0x00000000  0x00000000 
0x8d9d60:  0x00000000  0x00000000  0x00000000  0x00000000 

Answer 1

有趣的问题。答案是：在Linux（在这里我假设你跑你的程序）这样的：

(gdb) call calloc(1, 32) 

不会libc.so.6调用calloc。

宁可叫calloc从ld-linux.so.2。而那calloc是非常小的。它预计仅从ld-linux.so.2本身被调用，并且它假设它访问的任何页面都是“干净的”并且do not require a memset。 （也就是说，我无法使用glibc-2.19重现不清晰的calloc）。

可以证实这一点，像这样：

#include <stdlib.h> 
int main() 
{ 
    void *p = calloc(1, 10); 
    return p == 0; 
} 

gcc -g foo.c -m32 && gdb -q ./a.out 

Reading symbols from ./a.out...done. 
(gdb) start 
Temporary breakpoint 1 at 0x8048426: file foo.c, line 4. 
Starting program: /tmp/a.out 

Temporary breakpoint 1, main() at foo.c:4 
warning: Source file is more recent than executable. 
4   void *p = calloc(1, 10); 
(gdb) b __libc_calloc 
Breakpoint 2 at 0xf7e845a0 
(gdb) n 

Breakpoint 2, 0xf7e845a0 in calloc() from /lib32/libc.so.6 
(gdb) fin 
Run till exit from #0 0xf7e845a0 in calloc() from /lib32/libc.so.6 
0x0804843a in main() at foo.c:4 
4   void *p = calloc(1, 10); 

说明如何从程序calloc命中断点＃2呼叫。

(gdb) n 
5   return p == 0; 
(gdb) call calloc(1,32) 
$1 = 134524952 

注意，从上面的GDB调用没有不命中断点＃2。

让我们再试一次：

(gdb) info func calloc 
All functions matching regular expression "calloc": 

Non-debugging symbols: 
0x08048310 calloc@plt 
0xf7fdc820 calloc@plt 
0xf7ff16a0 calloc 
0xf7e25450 calloc@plt 
0xf7e845a0 __libc_calloc 
0xf7e845a0 calloc 

(gdb) info sym 0xf7ff16a0 
calloc in section .text of /lib/ld-linux.so.2 ## this is the wrong one! 

(gdb) break *0xf7ff16a0 
Breakpoint 3, 0xf7ff16a0 in calloc() from /lib/ld-linux.so.2 

(gdb) disable  
(gdb) start 
Temporary breakpoint 7 at 0x8048426: file foo.c, line 4. 
Starting program: /tmp/a.out 

Temporary breakpoint 7, main() at foo.c:4 
4   void *p = calloc(1, 10); 
(gdb) ena 3 
(gdb) n 
5   return p == 0; 

注意，断点＃3做上述不火（因为 “真正的” __libc_calloc叫）。

(gdb) call calloc(1,32) 

Breakpoint 3, 0xf7ff16a0 in calloc() from /lib/ld-linux.so.2 
The program being debugged stopped while in a function called from GDB. 
Evaluation of the expression containing the function 
(calloc) will be abandoned. 
When the function is done executing, GDB will silently stop. 
(gdb) bt 
#0 0xf7ff16a0 in calloc() from /lib/ld-linux.so.2 
#1 <function called from gdb> 
#2 main() at foo.c:5 

QED。

更新：

我没有看到 “信息FUNC释放calloc”

的输出LD-Linux版本，我认为你在info func看取决于你是否安装了调试符号。对于（64位）的glibc 与调试符号，这里是我所看到的：

(gdb) info func calloc 
All functions matching regular expression "calloc": 

File dl-minimal.c: 
void *calloc(size_t, size_t);   <<< this is the wrong one! 

File malloc.c: 
void *__libc_calloc(size_t, size_t); <<< this is the one you want! 

Non-debugging symbols: 
0x0000000000400440 calloc@plt 
0x00007ffff7ddaab0 calloc@plt 
0x00007ffff7a344e0 calloc@plt 

这里是另一种方式来弄清楚什么calloc GDB认为它应该叫：

(gdb) start 
Temporary breakpoint 1 at 0x8048426: file foo.c, line 4. 
Starting program: /tmp/a.out 

Temporary breakpoint 1, main() at foo.c:4 
warning: Source file is more recent than executable. 
4  void *p = calloc(1, 10); 
(gdb) p &calloc 
$1 = (<text variable, no debug info> *) 0xf7ff16a0 <calloc> 
(gdb) info sym 0xf7ff16a0 
calloc in section .text of /lib/ld-linux.so.2 

或者，为了完整性，使用带调试符号的64位glibc：

(gdb) start 
Temporary breakpoint 1 at 0x400555: file foo.c, line 4. 
Starting program: /tmp/a.out 

Temporary breakpoint 1, main() at foo.c:4 
4  void *p = calloc(1, 10); 
(gdb) p &calloc 
$1 = (void *(*)(size_t, size_t)) 0x7ffff7df1bc0 <calloc> 
(gdb) info sym 0x7ffff7df1bc0 
calloc in section .text of /lib64/ld-linux-x86-64.so.2