How is no operation used in buffer overflow?

How is no operation used in buffer overflow?

Basically no operation is used to make sure that our exploit doesn’t fail , because we won’t always point to the right address , so we add stuff that doesn’t do anything and we point to them , Then when the program executes it will reach those NOPs and keeps executing them (does nothing) until it reaches the shellcode.

How to find the address of the buffer?

If we type x/24wx $esp it will show us ( x/) 24 ( 24wx) words at the top of the stack ( $esp ). At the second line we see this address 0xbffff730 and it holds values of 0x41414141 and we already know that 41 is the hex of “A” which was our input to the program so we know that this address is where the buffer starts.

How to get buffer size with shellcode injection?

This time we don’t have a function to execute ,we have to find the address of the EIP and make it point to our “evil input” (shellcode), I will explain in a moment. Let’s take a break and take a quick look at another way to get the buffer size , I wanted to show this quickly because we have already done 50 % of it ..

How to find the EIP address of a buffer?

We can do it manually by calculating the distance between the buffer start address and the EIP address , We have already got the EIP address so let’s get the start of the buffer. If we type x/24wx $esp it will show us ( x/) 24 ( 24wx) words at the top of the stack ( $esp ).

Is the compiler responsible for buffer overflow in C + +?

But the problem with these functions is that it is the programmer responsibility to assert the size of the buffer, not the compiler. Every C/C++ coder or programmer must know the buffer overflow problem before they do the coding. A lot of bugs generated, in most cases can be exploited as a result of buffer overflow.

Why is 8 bytes too big for buffer overflow?

Input : 123456789 (9 bytes) “Segmentation fault” message will be displayed and the program terminates. The vulnerability exists because the buffer could be overflowed if the user input (argv [1]) bigger than 8 bytes. Why 8 bytes?