When using a digital pin as input, one has the option of using the internal arduino resistor, which is connected to VCC; that is setting the digital pin as INPUT_PULLUP, and then the pin reads HIGH. Does this uses electricity?.
I mean the resistor is connected directly to VCC and I presume electricity is constantly flowing through the system, no?. as oppose to having the pin as just INPUT?
Only the leakage current on the input itself is constantly consumed if there is no external connection. This is usually much less than the waste caused by having the input floating.
If a voltage between 0.8 V and 2 V is applied to the input for a prolonged period of time, this situation becomes critical and should not be ignored, especially with higher bit count and more dense packages [...]. For example, if an 18-bit transceiver has 36 I/O pins floating at the threshold, the current from VCC can be as high as 150 mA to 200 mA.
The pullups have a value between 20k and 50kOhm. If the input is connected to ground (button pressed), between 0.1mA and 0.25mA is flowing through the resistor. So bare anything.
See Deviation of power consumption between datasheet and real world? - there we established that leaving the pins input consumed considerably more power than input_pullup. That was with the pins not connected to anything.
I presume electricity is constantly flowing through the system- only if there is a load. With no load there is nowhere for it to flow, except as leakage current, as Ignacio Vazquez-Abrams said.