Okay, I can see the changes.
As shown now, the LEDs on the receiving IC will block the data high as they are reversed. The LEDs on the transmitting side aren't a good idea either. If you want LEDs you need to put them from the data line to ground through a resistor, not in series with the connection to cable socket.
Just seen your picture - it looks like you have removed the LEDs, good (I was just about to suggest this.. :-) )
So now it looks as if you have direct connections from IC1 to IC2. If this is the case then if the code (looks reasonable at a glance) and IC wiring are correct then it should work.
If you can confirm with a multimeter that the input pins are seeing a high (or low) voltage and the read value is different, then this would confirm the issue is one or the other of the above. Maybe just apply a known voltage directly and see if you can read that okay)
However, if you are reading different values when the pullups are on/off then that would seem to indicate the read is correct. Try reading a direct voltage and post results, I'm just checking the datasheet for the ICs, will add more shortly.
EDIT - about the pullups:
You can use the internal pullups if you don't mind the line "relaxing" to high when not driven (i.e. default state 1) These are often used for interfacing with open drain buses, or for button to ground, etc to save an external pullup.
If you want to have the lines default state low though, (as is the case for you) you need a pulldown to stop the high impedance floating. Since the IC in question doesn't have internal pulldowns, you need to add them externally.
EDIT - Doh! I've just seen the problem...
In your code you set 1 pin at a time to output and all the rest to inputs. This means that if you have internal pullups on, the undriven pins will default to high! When a pin is set to input, it is high impedance, so effectively it's like disconnecting that end of the line, and the weak pullups will pull the receiving end high.
You need to keep all pins as outputs, and just set one high at a time, this will keep all the pins driven - try this with the pullups on, it should work.
If you know the lines will be driven correctly all the time, you don't need the pullups, but it doesn't hurt to keep them on.
Here is the relevant code (in the intialise registers function):
i2c_start_wait(SLAVE_ADDRESS(0x4E)+I2C_WRITE); // Address Slave 1
i2c_write(0x00); // Set memory pointer to the IODIRA register (IODIRA address is 0x00 - see Page 9)
i2c_write(~(Value)); // Set only one pin at a time as an output and everything else as inputs
i2c_stop();
Change it to:
i2c_start_wait(SLAVE_ADDRESS(0x4E)+I2C_WRITE); // Address Slave 1
i2c_write(0x00); // Set memory pointer to the IODIRA register (IODIRA address is 0x00 - see Page 9)
i2c_write(0x00); // Set all pins as outputs
i2c_stop();
When in doubt, refactor.
It's not likely a hardware issue, but to rule it out, I would try replacing the C calls with the exact inline assembly instructions described in section 7 of the datasheet, following all of its recommendations (disabling interrupts, polling WR to test for doneness, etc.)
If you get it working, make your own function/library from the inline assembly and carry on.
Best Answer
Latch is a kind of memory of one bit.
Let's use the picture in manual:
When you write a bit in a I/O pin, you're storing this bit from Data Bus to the Data Register (D-FlipFlop). If TRISx of this bit is 0, so data from Q of the Data Register will be in the I/O pin. Write in LATx or PORTx is the same. See below in red:
On the other hand, read from LATx is different of read from PORTx.
When you're reading from LATx, you're reading what is in the Data Register (D-FlipFlop). See picture below in green:
And when you read from PORTx, you're reading the actual I/O pin value. See below in blue:
PIC uses read-modify-write to write operations and this can be a problem, so they use this shadow register to avoid it.