What Does Sink Eight TTL Inputs Mean in the Context of AT89C51's Port 0?

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I am trying to master "AT89C51 datasheet" by understanding it completely. But my limited knowledge is hindering me. help me to overcome all obstacles.

Port 0 is an 8-bit open-drain bi-directional I/O port. As an
output port, each pin can sink eight TTL inputs. When 1s
are written to port 0 pins, the pins can be used as highimpedance
inputs.

It says that Port 0 is an open-drain port. Does it mean that it can only sink current and can't source current? On the next line it says that "as an output port, each pin can sink eight TTL inputs." what does 'eight TTL input' mean here? Next it says that "When 1s re written to port 0 pins, the pins can be used as high impedance inputs." Does it mean that when the port 0 pins are set high, they won't sink any more current. They won't source any current either. all the current from the pull-up resistor will drive my circuit?
 
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snshusat161 said:
It says that Port 0 is an open-drain port. Does it mean that it can only sink current and can't source current?

yes. you need a pull-up resistor to make this source current.

On the next line it says that "as an output port, each pin can sink eight TTL inputs." what does 'eight TTL input' mean here?

it means you can connect it to 8 TTL inputs and it's not too big of a load.

Next it says that "When 1s re written to port 0 pins, the pins can be used as high impedance inputs." Does it mean that when the port 0 pins are set high, they won't sink any more current.

well, the shouldn't sink anything if they are "high". i am not sure if these pins double as inputs, but if they do, i would think you would need a bit to tell the chip that they're inputs instead of outputs. but, conceptually, if you want to use them as input, it makes sense that you write "1" to those bits as outputs, which turns them into a no-connection to ground ("0" virtually shorts the pin to ground with some ability to sink current) and then you could "read" the voltage on those open-drain pins which could be driven by something else.

They won't source any current either. all the current from the pull-up resistor will drive my circuit?

that's the idea of a pull-up resistor. to drive the rest of the devices connected to the pin when your open-collector or open-drain pin cannot supply positive output current to do it.