expressing a statement most compact method

Projects Programming
1

Hi
Is there a more compact method of expressing this statement below, I'm using it to shift the bits 1 position to the left ? bitIndex = bitIndex <<1;

// sample of code to reading data from a MT8870 DTMF ic

  const byte pins[] = {11,10,9,8};
  byte bitIndex = 1;
  byte val = 0;
  byte data = 0;
  //read the 4 data pins from MT8870
  for (byte i=0; i<4; i++)
    {
      val = digitalRead(pins[i]); 
      if (val)
        data |= bitIndex;  
      bitIndex = bitIndex << 1; //shift bits left   
    }

Thanks Don

2

Since HIGH is 1 and LOW is 0:

 const byte pins[] = {11,10,9,8};
  byte val = 0;
for (byte i=0; i<4; i++)
    {
      val |= digitalRead(pins[i]) << i; 
    }
3

Are you trying to make this specific statement more compact?

bitIndex = bitIndex << 1; //shift bits left

If so, a slightly more compact form would be:

bitIndex <<= 1; //shift bits left

If you're trying to make the overall for loop more compact then you could replace it with something like this:

for (byte i=0; i<sizeof(pins); i++)
{
    bitWrite(data, i, digitalRead(pins[i])); 
}
4

Thanks John, that's really simply. I struggle the most simple ways of coding my solutions to the task at hand

5

I'm looking to build a binary number of the input pins I listed.

6

Thanks Peter H too

7

Multiplying by 2 has exactly the same effect as shifting the bits one place left.

8

On the Arduino Uno this can be done with a single read:

  data = PINB & 15;

But that's just because pins 11-8 happen to be PB3-0.

9

Didn't think of that one I just happen to pick those pins.

Is there a problem solving in c book anyone can recommend to me so I
Can get more experience solving these type of problems. I feel like Im doing things the hard way kinda step by step. It's very interesting how different people code the same task.

Don

10 
const uint8_t pins[] = { 11, 10,  9,  8 };
for ( size_t i = sizeof(pins); i--; )
{
    val |= digitalRead(pins[i]) << 1; 
}
11

lloyddean:

const uint8_t pins[] = { 11, 10,  9,  8 };

for ( size_t i = sizeof(pins); i--; )
{
    val |= digitalRead(pins[i]) << 1;
}

That's not going to work. If you unroll the loop you get:

   val |= digitalRead(pins[3]) << 1;
   val |= digitalRead(pins[2]) << 1;
   val |= digitalRead(pins[1]) << 1;
   val |= digitalRead(pins[0]) << 1;

If all for pins are HIGH that is equivalent to:

   val |= 2;
   val |= 2;
   val |= 2;
   val |= 2;
12

A minor sidenote: My experince is that the compiler is not all that stupid - it usually does not matter how "compact" you write the code, it ends up with the same generated code. So unless one is having a timecritical code, write things the obvious human friendly way. (Reading tones from DTMF signal is probably not going to break if the loop takes 10us or 20us to do)

13

Yes I know by it got posed prematurely by accident then I lost my internet connection for the evening and I was unable to remove, correct it.

My apologies