This can be illustrated by an example,
#include "stdio.h"
int *ptr;
void func2()
{
int k = 300;
}
void func1()
{
int t = 100;
ptr = &t;
}
int main(int argc, char *argv)
{
func1();
printf("The value of t=%d\r\n",*ptr);
func2();
printf("The value of t=%d\r\n",*ptr);
}
On my machine, I got the following.
joshis1@(none) temp]$ ./ud.out
The value of t=100
The value of t=300
This tells that the value of t was not guaranteed. Once the scope of t was over, the stack space was allocated to k. Thus, ptr was accessing the same address - memory location . But the variable scope was over. You will get the consistent result, in case you don't call func2(); Thus, compiler doesn't guarantee the outcome -> This is called the undefined behaviour.
