Assuming that you want a regulated 5 volt source which can supply 1 ampere into a load, and that you want to use use a 7805 connected to a 9 volt supply to get that 5 volt output, the 7805 would have to dissipate: \$\ Pd =(Vin - Vout)\times Iload \ \ =\ \ (9V-5V) \times 1A = 4\ watts\$, while the load was dissipating \$\ 5V \times 1A = 5\ watts \$
The input power would, therefore, be 9 watts and, since efficiency = \$ Pout/Pin \$, we'd wind up with an efficiency of about 55%.
A switcher would run, typically, at about 80%, so it'd be dissipating 1.25 watts regardless of the input voltage - within limits, of course - and, compared to the 7805's 4 watts, would run much cooler.
The advantage of using a lower voltage on the input of a linear regulator is that the regulator will run cooler but with, say, 2 volts of headroom needed for a 7805 to run properly, it'll still be wasting 2 watts compared to the switcher's 1.25.
The downside for the switcher is that its output will be a little noisier then the linear's and it'll be more expensive. Maybe. By the time you get finished with the heat sink, the mounting hardware, and the the heat sink compound, who knows???