Why are we not allowed to return the address of the local variable from a function ?

[radha gogia]

int *g(void) 
{ 
int x = 10; 
return (&x); 
} 
 

what's the problem in returning the address of the local variable ?so even if its address of dead variable then still that memory location exists so it shuld give garbage value,


I am unable to get this one that how another function which uses the same local variable gets the same stack frame where the previous function foo resided ?

why is the op 57 here ?

#include<stdio.h> 
#include <stdlib.h> 
int * foo(){ 
int a = 5; 
return &a; 
} 
void boo(){ 
int a = 7;
 
} 
void boo1(){ 
int a = 57;
 
} 
int main(){ 
int * p = foo(); 
boo(); 
boo1(); 
printf("%dn",*p); 
}

[weaknessforcats]

This is not true:

I am unable to get this one that how another function which uses the same local variable gets the same stack frame where the previous function foo resided ?

Each function has its own stack frame. The stack frame exists until the function completes. So in your example, when function g returns, the stack frame for foo no longer exists. That means the local variable a no longer exists and if you return the address of a you provide your calling function with an address that maybe is invalid.

Unfortunately, the stack frame for a called function might actually exist for a while after the function completes. The calling function might be able to use the local variable briefly. This leads to a run time crash of heap corruption that is very difficult to debug.

In the case of a calling function passing the address of a local variable to a called function, that is OK. That's because the stack frame for the calling function will exist the entire time the called function is executing. However, this is done by using called function arguments.