Buffer Overflow Attack

Question

I'm trying to execute a very simple buffer overflow attack. I'm pretty much a newbie to this. So, if this question is stupid, please excuse me :-)

The code:

#include<stdio.h>
#include<stdlib.h>

int i, n;

void confused(int i) 
{
 printf("**Who called me? Why am I here?? *** %x\n ", i);
}

void shell_call(char *c) 
{
 printf(" ***Now calling \"%s\" shell command *** \n", c);
 system(c);
}

void victim_func()
{
 int a[4];
 printf("Enter n:  ");  scanf("%d",&n);
 printf("~~~~~~~~~~~~~ values and address of n locations ~~~~~~~~~~");
 for (i = 0;i <n ;i++) 
  printf ("\n a[%d] = %x, address = %x", i, a[i], &a[i]);
 printf("\nEnter %d HEX Values \n", n);

 // Buffer Overflow vulnerability HERE!

 for (i=0;i<n;i++)  scanf("%x",&a[i]);
   printf("Done reading junk numbers\n");
}

int main() 
{
 victim_func();
 printf(“\n done”);
 return 0; 
}

When I use objdump to get the function addresses, I have the following:

main(): 0x804854d
Address of main() where printf() is called: 0x8048563
victim_func(): 0x8048455
confused(): 0x8048414

Now, what I want is to jump to the function 'confused()' from victim_func() by overflowing the buffer there, and overwriting the return address to the address of confused(). And I want to return back from confused() to the printf() statement in main, and exit normally. So, I provide the following input

Enter n: 7
Enter 7 HEX values:
1
2
3
4
5
8048414 (This is to jump to confused)
8048563 (this is to jump to printf() in main)

Although, the program prints "Done" from that printf statement, it is jumping back to victim_func() and prints "Enter n:"

What am I doing wrong? Any help would be greatly appreciated!

PS: I'm not sure if I have put the question right. Please let me know, if any more information is needed.