why does returning the local variable address throws an error but returning the local variable value don't?

Question

I can not understand, what is the difference between following function bodies

int func(void){
    int A = 20;
    return A;
}

and

int* func(void){
    int A = 20;
    return &A;
}

why returning the values does not throw the error of the segmentation fault but returning the address do?

Answer 1

A local variable with automatic storage duration is not alive after exiting the scope of the function where it is defined. So the returned pointer will be invalid and de-referencing such a pointer invokes undefined behavior.

If you will change the second function the following way

int* func(void){
    static int A = 20;
    return &A;
}

then returning a pointer to the variable A will be correct because the variable having static storage duration will be alive after exiting the function.

A returned value can be assigned to a variable or discarded. So there is nothing wrong.

Answer 2

Attempting to target the more specific question asked in follow-up comments on several other answers:

if the address of the local variable after the scope gets deleted then how the value of that address preserved? in case of returning local variable value from the function? this is I can not understand

Returning a value makes a copy of that value, into a storage location provided by the caller. You presumably have something like

void call_func(void) {
   int n = func();
   printf("%d\n", n);
}

as the function that calls func -- so, when func returns the value of A, that value gets copied into n.

Returning a pointer to A is actually exactly the same: a value is copied into a storage location provided by the caller. Only now the value that gets copied, is the address of the storage location formerly occupied by A.