What does sizeof do?

To add to Peter Alexander's answer: sizeof yields the size of a value or type in multiples of the size of a char---char being defined as the smallest unit of memory addressable (by C or C++) for a given architecture (and, in C++ at least, at least 8 bits in size according to the standard). This is what's generally meant by "bytes" (smallest addressable unit for a given architecture) but it never hurts to clarify, and there are occasionally questions about the variability of sizeof (char), which is of course always 1.

sizeof(x) returns the amount of memory (in bytes) that the variable or type x occupies. It has nothing to do with the value of the variable.

For example, if you have an array of some arbitrary type T then the distance between elements of that array is exactly sizeof(T).

int a[10];
assert(&(a[0]) + sizeof(int) == &(a[1]));

When used on a variable, it is equivalent to using it on the type of that variable:

T x;
assert(sizeof(T) == sizeof(x));

As a rule-of-thumb, it is best to use the variable name where possible, just in case the type changes:

int x;
std::cout << "x uses " << sizeof(x) << " bytes." << std::endl
// If x is changed to a char, then the statement doesn't need to be changed.
// If we used sizeof(int) instead, we would need to change 2 lines of code
// instead of one.

When used on user-defined types, sizeof still returns the amount of memory used by instances of that type, but it's worth pointing out that this does not necessary equal the sum of its members.

struct Foo { int a; char b; };

While sizeof(int) + sizeof(char) is typically 5, on many machines, sizeof(Foo) may be 8 because the compiler needs to pad out the structure so that it lies on 4 byte boundaries. This is not always the case, and it's quite possible that on your machine sizeof(Foo) will be 5, but you can't depend on it.