& vs * in param list?

What the difference between

Code:

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void myfunc(mytype &a);
void myfunc(mytype *a);

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From what I've found reading online, in both cases, modifying the 'a' modifies the param passed into the function after it returns. But if mytpe is a struct, in the 1st, it requires . notation and the 2nd needs -> notation.

I know only a pointer to a is passed in the 2nd, so not much gets added on the stack when the func is called, but what about the 1st? Is pointers on my system are 4 bytes, is their only an extra 4 bytes added to the stack in the 1st?

Hopefully that q was clear.

Yeah, references are passed as pointers under the hood.

Quote:

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Originally Posted by King Mir

Yeah, references are passed as pointers under the hood.

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No, that's not guaranteed.
A reference is an alias. If you pass "b" to a function which takes a reference named "a", then "a" is an alias for "b". And that is all.
So whatever happens to a, happens to b.

Above you have both the real-world answer and the theoretical answer. Great huh!
Neither necessarily uses stack space. Registers are often used to pass parameters. Stack space may not be required for them at all if the function is simple enough.

What you also should know is that using a reference means that:
you can't accidentally reassign 'a' to point to something else, and
the passed in parameter cannot be NULL.

Quote:

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Originally Posted by Elysia

No, that's not guaranteed.
A reference is an alias. If you pass "b" to a function which takes a reference named "a", then "a" is an alias for "b". And that is all.
So whatever happens to a, happens to b.

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By the same token, pointers aren't guaranteed to be memory addresses either. What's under the hood is be definition implementation defined.

But in the common case, pointers and references are passed through the stack as 4 byte memory addresses. And both have the same overhead.