A focal length is a physical property of a lens.
The field of view provided by that lens in 35mm (aka 135-format) nomenclature is 1.6x the marked focal length (whether it be an EF-S lens with a 'shorter' marked focal length, or an EF lens designed for full-frame cameras, with whatever focal length it specifies.
EFOV = equivalent field of view.
An 18mm lens is an 18mm lens, whether it's EF or EF-S.
The '-S' in EF-S designates 'short back-focus', meaning that the lens's rear element protrudes further into the chamber, and is thus closer to the sensor.
Canon designed the EF-S system such that the bayonet mount would also accommodate EF lenses (which are themselves designed for 135-format cameras).
The difference with an EF lens is that the sensor cannot 'see' the entire imaging circle, whereas an EF-S lens is designed such that the the imaging circle's diameter is the same size as the diagonal length of the sensor.
Greco is correct. What he means is that an 18mm EF-S lens on an APS-C DSLR camera will provide the equivalent framing of a 28mm lens on a 35mm camera (or 'full-frame' DSLR).
In short:
- Canon EOS 60D + 18mm EF-S lens = equivalent of 28.8mm framing on a 35mm camera.
- Canon EOS 60D + 18mm EF lens = equivalent of 28.8mm framing on a 35mm camera.
- Canon EOS 5D + 28mm EF lens = very close framing to the above two examples.
- Canon EOS 60D + 28mm EF lens = equivalent of 44.8mm framing on a 35mm camera.
I hope this helps.
As to whether or not Grego's explanation was helpful to the person whose query he answered, I cannot say.
What I can say is that I understand this stuff, but it can be confusing to those who don't.
It makes less sense to people who don't know what kind of framing various lenses provide on a 35mm camera; it's another language, and dare I say it, for the beginners, no matter what format of camera they have, most of them just want to know how much 'stuff' is in the scene at the wide end of a lens, and how much it 'zooms' (ie, its magnification factor).