Confusion in Scope and Life Time of a local variable in c/c++ -

this question has answer here:

• is undefined behavior dereference dangling pointer? 4 answers

my question when lifetime of local variable @ block level why pointer still printing value of local variable outside block

 #include<iostream>    using namespace std;  int main(){      int *p;     {         int n;         n=5;         p=&n;     }      cout<<*p;     return 0; } 

scope refers the availability of identifier.
life time refers actual duration object alive , accessible legally during execution of program. distinct things.

your code has undefined behaviour because lifetime of object n on @ closing } because access through pointer.

a simple example might make clearer:

#include<stdio.h> int *func() { static int var = 42; return &r; }  int main(void) { int *p = func(); *p = 75; // valid. } 

here, var has static storage duration. i.e. it's alive until program termination. however, scope of variable var limited function func(). var can accessed outside func() through pointer. valid.

compare program. n has automatic storage duration , lifetime , scope both limited enclosing brackets { }. it's invalid access n using pointer.

however, if make change (n) storage class static can object alive outside enclosing brackets.