Question

I am trying to write a function in C that will shift out the individual bits of a byte based on a clock signal. So far I have come up with this…

void ShiftOutByte (char Data)
{
    int Mask = 1;
    int Bit = 0;
    while(Bit < 8)
    {
        while(ClkPin == LOW);
        DataPin = Data && Mask;
        Mask = Mask * 2;
        Bit++;
    }
}

where DataPin represents the port pin that I want to shift data out on and ClkPin is the clock port pin.

I want the device to shift out 8 bits, starting on the LSB of the byte. For some reason my output pin stays high all the time. I am certain that the port pins are configured properly so it is purely a logical issue.

Answer 1

Your solution is almost there, but there a few problems:

• When clock goes high, you will pump out data in a burst before the clock has a chance to go low again, so you need to pause while clock is high before you check for it being low (unless the hardware pauses execution until clock goes low). This can be done either at the end of the loop, or at the beginning. I chose the beginning in the sample below, because it allows you to return from the function while clock is still high, and do some processing while it is still high.
• You used the logical and (&&) instead of bitwise and (&) as others have pointed out
• I am not familiar with your architecture, so I can’t say if DataPin can only accept 0 or 1. But that is also another point where things may go wrong. Data & Mask will return a bit that is shifted left (1, 2, 4, 8, …) depending on the bit position.

So, in summary, I would do something like:

void ShiftOutByte (unsigned char Data)
{
    int Bit;
    for(Bit=0; Bit < 8; ++Bit)
    {
        while(ClkPin == HIGH);
        while(ClkPin == LOW);
        DataPin = Data & 1;
        Data >>= 1;
    }
}

Note: I have used a for loop instead of the while ... Bit++ pattern for clarity of purpose.