运行了strtok后的内存如何释放？

dyte200

程序：
[php]
#include <string.h>
int
main ()
{
  char *a = NULL;
  char *b = NULL;
  a = strdup ("hello world !");

  b = strtok (a, " ");

  printf ("a=%s\n", a);
  printf ("b=%s\n", b);
  printf ("\n");

  b = strtok (NULL, " ");

  printf ("a=%s\n", a);
  printf ("b=%s\n", b);
  printf ("\n");

//free(b);
  free (a);

  printf ("a=%s\n", a);
  printf ("b=%s\n", b);
}

[/php]
结果显示：
a=hello
b=hello

a=hello
b=world

a=
b=world
怎样将原string完全释放掉？

x11

看来strdup后你得自己保存一份这个指针值了，呵呵

zhllg

已经释放了
释放 不等于 把整个那块区域以0填充
目前glibc的实现只是把前4个字节弄成0
试试这个

1. 
   
2. #include <string.h>
   
3. #include <stdio.h>
   
4. #include <stdlib.h>
   
5. int main ()
   
6. {
   
7.   char *a = NULL;
   
8.   char *b = NULL;
   
9.   a = strdup ("hello world !");
   
10. 
    
11.   b = strtok (a, " ");
    
12. 
    
13.   printf ("a=%s\n", a);
    
14.   printf ("b=%s\n", b);
    
15.   printf ("\n");
    
16. 
    
17.   b = strtok (NULL, " ");
    
18. 
    
19.   printf ("a=%s\n", a);
    
20.   printf ("b=%s\n", b);
    
21.   printf ("\n");
    
22. 
    
23. //free(b);
    
24.   free(a);
    
25. 
    
26.   printf ("a=%s\n", a);
    
27.   printf ("b=%s\n", b);
    
28.   printf ("\n");
    
29. 
    
30.   a = (char *)malloc(sizeof("hello world !") + 1);
    
31.   printf("now malloc an buffer whose size is equal to "hello world !"\n"
    
32.          "this buffer is just the buffer which strdup previously returned\n"
    
33.          "which means strdup returned buffer is reused, coz it's been freed\n");
    
34.   strcpy(a, "hello earth !");
    
35.   printf("now copy "earth" to where "world" originally resides\n");
    
36.   printf ("and b becomes: %s\n", b);
    
37. 
    
38. }
    

复制代码

x11

哦，对了，指针a的值并没有被改变
free(a)就释放了
前面我想错了，呵呵