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

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

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?

Solution

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.