stack frame analysis of programs in Linux and modifying function addresses

Here's a piece of code:

#include <stdio.h> #include <unistd.h> #include <stdlib.h>void fun () {printf ("I am the Evil func\n"); Exit (1);}    int fun1 (int a,int b) {int *p=&a;    p--;    *p=fun;    int C=0XCCCC; return c;} int main () {printf ("Begin run..    \ n ");    int a=0xaaaa;    int b=0xbbbb;    Fun1 (A, b);    printf ("Right End");    return 0; }

Execution results under Linux:

Main call FUN1 Call fun---exit exit

For the problem with this result:

Obviously did not call the fun function, why do the fun function, the function is how to jump?

How does the stack frame save the information and return it?

Analysis Reason:

View important stack segments in assembly code

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Simple stack frame diagram

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The following program accesses the value of the variable y through a pointer, not the variable name y

#include <stdio.h>int fun (int x,int y) {int *p = x;    p--; return *p;}    int main () {int x=1;    int y=2;    int Ret=fun (x, y);    printf ("Y's value is%d\n" ret); return 0;}

The principle is that after p-->x,p--, p points to Y

At this point *p is accessing the value of Y.

This article from "Momo is spicy moe" blog, please be sure to keep this source http://momo462.blog.51cto.com/10138434/1791894

stack frame analysis of programs in Linux and modifying function addresses