By "no resistance" I'm going to assume you mean "zero resistance" and say ALL of the current will flow through the zero resistance path. Now, in the real world, there will likely be some resistance (unless the shunt is a superconductor), in which case the current will divide between the two.
So, there are several modes to look at:
1. diode forward biased.
2. diode reverse biased but not in avalanche or breakdown.
3. diode in avalanche or breakdown.
For case 2, the if any current flows through the diode it will be the reverse leakage current. But, if the "resistor" has 0 ohms, then, still, all of the current will flow through that 0 ohm path. No current will flow through the diode. Otherwise, the current will divide between the two. In most cases the current through the diode will be negligible.
For case 3, the diode will have a fairly constant [reverse] voltage across it and the current will be hard to predict (in the case of a voltage source connected directly across the resistor diode pair).
For case 1, there will also be a fairly constant [forward] voltage across the diode--not as constant as in case 3, though. Again, the current will be hard to predict.
What will happen if a voltage source is placed directly across the resistor diode pair in case 1 and case3? The current through the diode will rise to a limit determined by the internal resistance of the source -- or it will generate lovely curls of acrid smoke (in the case that the voltage source has a low enough internal resistance to allow enough current to flow to fry the diode ;)
Because the diode is a non-linear device with characteristics that vary with temperature (much more so than most resistive materials) and that vary between parts (no two are exactly alike -- of course that can be said for some resistive devices, too) -- it's a little difficult to come up with precise currents, especially since the math is a bit hairy:
http://en.wikipedia.org/wiki/Diode_modellingBasically the circuit becomes a diode with a resistor across it and another resistor (the internal resistance of the voltage source) in series with the resistor/diode pair.
For case 1, if you approximate the diode voltage at somewhere between .6 to .7 for small currents and upwards of 1 to 1.5 volts for large currents, then knowing that the voltage across all parallel circuit components is the same, the resistor will have the same voltage across it as the diode. So, you can simply use ohm's law to determine the current in the resistor [Ir = Vd/R], and the current in the diode will be the total current minus the current in the resistor. The total current can be determined with the following formula:
It = (Vs - Vd)/Rs
where *It* is the total current, *Vs* is the magnitude of the source voltage (measured with no load attached), *Vd* is the forward voltage of the diode, and *Rs* is the internal resistance of the voltage source.
If the voltage source is perfect (i.e. has an internal resistance of 0) and if the diode is capable of handling any amount of current, then what will happen when you put the resistor/diode pair across the voltage source is this: the forward voltage of the diode will rise until it is equal with that of the voltage source. If the magnitude of the voltage source is much more than a volt or two, there will be a massive amount of current flowing through that diode -- image how much current would have to be flowing in a diode for the forward voltage to be at, say, 5 volts -- remember, it's an exponential relationship!
Case 3 is similar to case 1. The diode will have it's avalanche or breakdown voltage across it, and the current will rise until limited by the source resistance. The parallel resistor will have the same voltage across it and ohms law applies.
