In the eyes of the application program ape, the C-language method of writing memory, as follows:
struct mystr * PTR;
ptr = malloc (sizeof (*PTR));
Ptr->filda= ' abc ';
ptr->fildb=89;
However, if it is a system program, no malloc, even need to design your own malloc, how to write memory?
Below for everyone to show a technology, this technology is more bloody and violent, not suitable for children.
Children over the age of 18 can continue to watch.
1 #include
2 #include
3
4 struct STR1 {
5 int i;
6 Char J;
7};
8
9 int main () {
struct str1 * PTR1;
One ptr1 = malloc (sizeof (*PTR1));
((int*) ptr1) = 23;
((char*) ((int*) PTR1 + 1)) = ' a ';
14
printf ("Now Ptr1->i Is:%dn", ptr1->i);
printf ("Now Ptr1->j Is:%dn", ptr1->j);
17
return 0;
19}
A daring little boy's shoes guess, ptr1->i is how much?
Yes, the output is this:
[Root@localhost ~]#./memwrite
Now Ptr1->i is:23
So, how much is ptr1->j?
If you know the answer, then why would he be the result?
If you haven't read it, add my QQ number.
If you understand the core principles of this hack, it's easy to read the Linux kernel's paging management, swap technology, etc.