It's all just indirection: The ability to not deal with data, but say "I'll direct you to some data, over there". You have the same concept in Java and C#, but only in reference format.
The key differences are that references are effectively immutable signposts - they always point to something. This is useful, and easy to understand, but less flexible than the C pointer model. C pointers are signposts that you can happily rewrite. You know that the string you're looking for is next door to the string being pointed at? Well, just slightly alter the signpost.
This couples well with C's "close to the bone, low level knowledge required" approach. We know that a char* foo consists of a set of characters beginning at the location pointed to by the foo signpost. If we also know that the string is at least 10 characters long, we can change the signpost to (foo + 5) to point at then same string, but start half the length in.
This flexibility is useful when you know what you're doing, and death if you don't (where "know" is more than just "know the language", it's "know the exact state of the program"). Get it wrong, and your signpost is directing you off the edge of a cliff. References don't let you fiddle, so you're much more confident that you can follow them without risk (especially when coupled with rules like "A referenced object will never disappear", as in most Garbage collected languages).
For the language difference (keeping only the function declarations below, since that's what's important only)
void execute( void (&func)() );
void g();
int main() {
void (*fp)() = g;
execute(fp); // doesn't work
execute(&g); // doesn't work either
execute(g); // works
}
It doesn't work, because it wants a function, not a function pointer. For the same reason that array answer rejects a pointer, this rejects a pointer too. You have to pass "g" directly.
For templates, it matters too
template<typename T>
void execute(T &t) { T u = t; u(); }
template<typename T>
void execute(T t) { T u = t; u(); }
Those two are very different from one another. If you call it with execute(g); like above, then the first will try to declare a function and initialize it with t (reference to g). The generated function would look like this
void execute(void(&t)()) { void u() = t; u(); }
Now you can initialize references and pointers to functions, but of course not functions itself. In the second definition, T will be deduced to a function pointer type by template argument deduction, and passing a function will convert it to that pointer parameter type implicitly. So everything will go fine.
I don't know why MSVC treats them differently for inlining - but i also suspect it's because function references appear more seldom.
Best Answer
Theoretically, they could be implemented in different ways.
In practice, every compiler I've seen compiles pointers and references to the same machine code. The distinction is entirely at the language level.
But, like cdiggins says, you shouldn't depend on that generalization until you've verified it's true for your compiler and platform.