To further complicate the answer above and answer a couple of other items at a fundamental level...
As mentioned the current in a circuit is shared by all devices in series connections (and divided amongst parallel ones), however a power source (battery in this case) can supply a variable current, depending on the load's "draw" from it. So the same battery can supply 1mA, or 1A, or anything in between (within it's fundamental range).
However, a circuit must always drop the voltage differential from the power source (not a variable).... so if a battery is a 7.2V battery, 7.2V of voltage times XXXmA of current is going to be dropped somewhere in the circuit. This gives you your power consumption in watts (W=VA or P=EI is technically the formula, but I'll use the W=VA).
How does this apply to the boost circuit, well, first, the boost circuit needs some voltage and current simply to operate, usually the "chip" involved has several transistors, each drops 0.6V individually (depending on the circuit, some power others, won't go into that in detail). Some part of the regulator is going to have to "see" the current output, so if the output is a 1A output, then at least one transistor or a combination of them is seeing 0.6V*1A, or 0.6 watts. So that is already lost as a given to start with (and quite a bit more). Other losses in the circuits will add up to the efficiency, say 85%. Of that, most is heat, but it's not "intentionally" lost here (unlike a linear regulator), it's just a function of operation.
Now the boost circuit is intended obviously to boost something, current, voltage, whatever... but you can't violate the fundamental law of V=IR in the circuit. Now let's say you have 3*3.3V LEDs running at the 1A desired rate. Obviously we'd need more voltage than the 7.2V source can provide to reach the 9.9V required for the LEDs to operate at 1A, but the battery cannot provide more than that, so instead what happens is the regulator stores available voltage in 2 source circuits (using inductors and capacitors) and as the available voltage (stored as potential energy) in one falls, the other is used and the first is recharged. Obviously to keep 2 sources charging and discharging to provide a greater voltage requires more total energy, and since you can't change the battery voltage, the current requirement is higher to makeup the difference (R doesn't change significantly in this example, although both inductors and capacitors do impose some resistance losses in operation).
That's probably clear as mud and some technical minor inaccuracies, but look at it another way, if the regulator is 100% efficient, then V=IR still applies. You are looking at 9.9V at 1A, at whatever resistance (R). Keeping R constant, if your source is 7.2V, then you have 9.9=1R and 7.2=XR.... 9.9/7.2=1.375 increase in voltage, to achieve that, you'll have to draw (in a 100% efficient battery at different rates), 1.375 more current, or in this case 1.375 amps (to get 1 amp output at 9.9V). If the convertor is 90% efficient, and the battery loses 5% efficiency due to increased draw (or is 95% of the 1A rating), then you get 1.375/(.9*.95) = 1.6amps of draw on the batteries.
Now of course that looks like a "constant voltage" scenario in my write-up, but in this case for simplicity I focused on the voltage numbers, as the LEDs require 9.9V to operate at 1A (so to supply a constant 1A current to the LED, the regulator will have to keep the voltage at 9.9V, if the voltage of the LED shifts, the voltage supplied by the regulator will shift to maintain the 1A current, but I didn't want to get into all that).
The advantage of a boost circuit is obviously 2-fold, getting the higher voltages needed to operate an LED at a high current, and current regulation to keep it constant.
A third advantage can also be a drawback if not limited, that the regulator will keep boosting the voltage as long as it can, drawing more and more current from the batteries as they deplete (as the voltage differential increases), until such point the batteries can't supply the current necessary to overcome the voltage differential, at which point your light suddenly dims and goes out, because the batteries are simply kaput (potentially damaging the batteries in the process as they get excessively undervolted).
