Is it good idea to return an address or a reference of a local variable True False?

Once the function returns it does not exist anymore and hence you should not return the address of a local variable. In other words the lifetime of a is within the scope( { , } ) of the function and if you return a pointer to it what you have is a pointer pointing to some memory which is not valid.

Can we return local variable in C?

The preferable way would be to use malloc() to reserve non-local memory. the danger here is that you have to deallocate ( free() ) everything you allocated using malloc() , and if you forget, you create a memory leak. For foo1() , you return a copy of the local variable, not the local variable itself.

Can a function return an address?

Because return by address just copies an address from the function to the caller, return by address is fast. The end result is that the caller ends up with the address of non-allocated memory (a dangling pointer), which will cause problems if used.

What happens if we return address of a local variable and solution?

Since the problem is that the return address is actually of a local variable, the solution would be to dynamically allocate memory to a variable to hold the address and pass it to the function as a parameter. Now, the variable is declared outside the function block and is no longer a local variable.

Should you always pass by reference C++?

if you want to change the stack object you are passing in, do so by ref. if you dont, pass it by value. if you dont wanna change it, pass it as const-ref. the optimization that comes with pass-by-value should not matter since you gain other things when passing as ref.

Why is passing by reference faster?

As a rule of thumb, passing by reference or pointer is typically faster than passing by value, if the amount of data passed by value is larger than the size of a pointer. Of course, if your called function needs to modify the data, your decision is already made for you…you need to pass by reference or pointer.

When a function call returns what happens to variables in the local scope?

A local variable retains its value until the next time the function is called A local variable becomes undefined after the function call completes The local variable can be used outside the function any time after the function call completes.

What happens to a local variable when a function call completes?

When the execution of the function terminates (returns), the local variables are destroyed. So it is not possible for a function to set some local variable to a value, complete its execution, and then when it is called again next time, recover the local variable.

What happens to a local variable when a function returns?

After a function returns, its stack frame is obliterated and local variables no longer exist. Therefore, a reference to a local variable (which is basically a pointer) would point into garbage memory on the stack.

Why can’t I return a reference to a local variable?

When you hang onto a reference to this deleted memory and attempt to use it, terrible things happen. This may crash or worse. The reason you can’t return a reference to a local variable is that it ceases to be valid after the function exits, much like the memory behind a pointer ceases to be valid after you delete the pointer.

When should I Be careful when returning an address from a variable?

You should always be careful when returning addresses to local variables; as a rule, you could say that you never should. static variables are a whole different case though, which is being discussed in this thread.

Is it possible to return a pointer to a local variable?

Therefore, a reference to a local variable (which is basically a pointer) would point into garbage memory on the stack. It is possible to return a pointer to a local variable, but a good compiler will warn you, and actually doing it is likely to trigger undefined behavior.