Why doesnt this XOR give output

Tie one of the gates to ground or supply with 5k resistor.

But is that how it's supposed to work in real situations? I tries this in another simulator, iCircuit in which this same circuit works like I expected.

Bad practice to leave unused gates floating all gates should be tailed to their respective logic. Otherwise if left to float they are susceptible to external noise or their own gate seepage with unpredictable results. Most simulators assume perfect part's. And often circuit's designed on simulators won't work in real life. This is where experience comes in handy. Simulation is only intended to provide demonstration of principle and the final real world design may in many ways differ substantially. Even the very complex simulation software like the spice suit don't get it right all the time.

Thanks a lot. It's been quite a few years since I studied electronics and I've pretty much forgotten most of the things

It is incomplete. Check mine "xor tester 2" #1 is with inverting xor try both

Listen to pip, hes one of the smartest people ive met so far and i havent even met him

If the other input is ¨open¨, the gate can decide on its own which logic level it has. To fix this, you either need a switch which switches between + 5V and ground or you need to use ¨pull up¨ or ¨pull down¨ resistors to make the input pin with the open conrtact reach a defined logic level. The 74LS series has built in pull ups. An open input pin will always be recognized as a logic ¨HIGH¨. Most CMOS chips could either choose a random state or even oscillate violently. In fact using a 74LS with open input pins is no good either since if one gate of a chip is switching, it causes a temporal short circuit whicxh can make open inputs be recognized as a ¨LOW¨ for a split second. So always add small capacitors (100nF) between tghe supply pins (and as close to the chip as possible) and connect all unused input pins (even of unused gates) to plus, ground any neighbour input or output.

Use pull-down resistors on input.

Put an inverter one to + and the other to ground then you would have a digital input.