Can u give the Solution

Hello Gurus,

Can u give any solution for the below code.


Thanks
Sandy




void change()
{
//Do any thing that will print value of i in main as 5;
//without changing code of main
}

int main()
{
int i = 5 ;

clrscr();

change();

i = 10;

printf("Value of i=%d",i);

getch();

return 1;
}

Code:

---

void change()
{
        int x = 0;
       
        printf("Value of i=%x\n",*((&x) + 22));
}

---

Visual C++ 6.0, standard Debug console settings.

Not portable, not useful.


Pete

TheMachineElf: That does cause i=5 to be printed in the function change(), but i is still set to 10 and that value is printed when execution returns to main(). I think the point of the question was to write change() in such a way so that i=5 is printed from the existing printf() statement in main(). SandyG, I doubt that there is any graceful, portable way to do this, and I can't imagine why you'd want one, but here's a terrible hack that does the job, at least in VC++ 6.0 with the Debug settings.

Code:

---

void change()
{
    __asm
    {
        mov eax,dword ptr [ebp+4]
        add eax,7
        mov dword ptr [ebp+4],eax
    }
}

---

It loads the return address that was stored during the call to change() and alters it so that once change() is finished, execution returns to the line after the assignment i = 10, so that line is never executed. Again, this is a terrible hack and will break if you even look at it the wrong way. It's good for nothing except as a curiosity. I can't think of any reliable way to do what you want.

maybe somehow cheat stdout to catch printed string and replace it with required?

This works on VC++ 7.1. Main still doesn't print out 5, but it doesn't print out 10.

Code:

---

void change()
{
        printf("Value of i=%d",5);
        stdout->_file = NULL;
}

---

Could also do this, but I guess technically it does change the code in main...

Code:

---

void change()
{
#define printf(x,y) printf(x,y/2)
}

---