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What is the memory address of my code?

Hi I am trying to find the memory address of where my data gets stored.

This is the assembly code of the function.

   0x08048b20 <+0>:     push   %ebp
   0x08048b21 <+1>:     mov    %esp,%ebp
   0x08048b23 <+3>:     sub    $0x28,%esp
   0x08048b26 <+6>:     lea    -0x14(%ebp),%eax
   0x08048b29 <+9>:     mov    %eax,(%esp)
   0x08048b2c <+12>:    call   0x8048990 <Gets>
   0x08048b31 <+17>:    mov    $0x1,%eax
   0x08048b36 <+22>:    leave  
   0x08048b37 <+23>:    ret   

My data which is a string gets stored at -0x14(%ebp) - (pretty positive). And I know the return address should be at 4(%ebp). What I am trying to do is set the return address to point at my code. And I can't use assembly to do this. I need to know the exact memory location of my where my function starts (which I know it stored at -0x14(%ebp)). Can anyone help me on how to get the memory location of that position?

The address on this line

0x08048b20 <+0>:     push   %ebp

The 0x08048b20 is the memory location for the function I am in, not %ebp right? My thinking is that if I get the memory location for %ebp, I can calculate the memory location of where my function starts.

I am using gdb, and I don't know how to get the memory address of it? Please, any help would be appreciated. Thank you.

EDIT: When I am standing at line <+6> in the assembly code, I print the values of %ebp and %esp, and they have an exact difference of 0x28 like they should. So I assumed that the address of %ebp is the value I get when I print it in gdb.

However, when I subtract 0x14 from this value (0xbfffb5d8) to give 0xbfffb5c4; it doesn't know where to jump to. If I have stored my string (which is actually byte code of disassembled code) at -0x14(%ebp), and I want to run that code; shouldn't I be putting the return address to -0x14(%ebp) or am I thinking completely wrongly?

Put a label before the first line of your function. Then referencing the label will give you the address of the first opcode in your compiled function. Unfortunately I don't understand your question enough to help you any more than that.

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