The other problem with a fixed resistor is that Vf falls as Tj rises.
This means that the current tends to increase as the device gets hotter, which increases the current, which makes it hotter...
This is the normal thermal runaway issue that you get when powering a LED from a constant voltage.
However the same thing can happen with a resistor. The issue is, will it run away?
(edit: runaway means increasing heat --> increasing power dissipation. Even without this, the power dissipation can cause the temperature to rise above the limit where the device fails, but that is a thermal failure, not a thermal runaway.)
Let's model the LED as a resistor.
If the LED current is nominally 1A, and the battery voltage is 12V and it has a 1 ohm resistor in series with it, then it must be acting as if it has a resistance of 11 ohms.
So by simple calculation, it must be dissipating 11W (I'm using simple figures, so ignore the excess dissipation)
Lets say that it gets heated by this and the internal resistance falls to 10.5 ohms
The current rises to 1.043A, and the power dissipation rises to 11.4W.
So now it's going to get even hotter...
Let's instead, assume that the series resistor is 13 ohms and the current is 1A and the LED appears to have a resistance of 11 ohms. (Which means our power supply must be 24V).
The LED sees the same power, so its resistance drops, lets say to 10.5 ohms again. This time the current rises to 1.021 A, giving a power dissipation of 10.95W -- The power falls! So the LED will cool slightly.
In the first case, the LED current can still run away, in the second it reaches an equilibrium quickly (actually it will reach equilibrium in the first case too, but at a current dramatically higher than the nominal current.
In these 2 cases, we have assumed a voltage source with zero internal resistance and a series resistor. This is exactly the same as a real (imperfect) voltage source with an internal resistance.
It's pretty clear that increased internal resistance is a good thing in this case because it stabalises the current.
Wouldn't it be great if we could have an infinite internal resistance? Impossible?
Enter the current source...
A perfect current source is an infinite voltage in series with an infinite resistance. Because the load resistance is always low compared with the impedance of the supply, the current does not change as the load resistance changes.
In practice a constant current source employs neither infinite voltages not infinite resistances (and with it infinite power dissipation), but employs some circuit with feedback to ensure a relatively constant current over some range of load resistances.
So we have the LED with an assumed resistance of 11 ohms connected to a 1A constant current source. The current is clearly 1A, and the power dissipated is 11W, as it heats up and the resistance falls to 10.5 ohms, the current remains the same and the power drops to 10.5 watts, which is a lot lower. The device can still get hotter, and can still exceed Tj(max) but since the current does not increase, and the power dissipated does not increase, failure will be from excess temperature caused by the nominal current, not by thermal runaway caused by ever increasing current.
If you use resistors, you can't exclude the possibility of thermal runaway, but you can ensure that device power falls as current increases. To do that, at least half the supply voltage must be dropped across the series resistor. Clearly for high currents (i.e. high power LEDs) this is extremely wasteful.
Things are actually a little worse than I've suggested above. LEDs are not resistors. Their resistance does not fall as their temperature rises, their Vf does. Vf is also dependant on current, so in the three cases above, if the second case reflected what actually happened, the first would be worse (more current, higher power), and the last would be better (lower Vf even lower power).
If you've read my tome on driving LEDs, you'll note that I recommend constant current sources for high power LEDs. The above is an explanation that I've not put in there which justifies that recommendation.