String to Binary/Byte for 595 Shift Register

Good afternoon folks. I have scrubbed google and I just can't seem to find the answer I am looking for.

I am using a 595 shift register to eluminate LED's (original huh?) . I am basically building a string that I can pass using shiftout. The string looks like this: "B01001000". In this example, I want to power the LEDS on pin 2 and 5.

My programming question is: How can I convert this String to a byte so that I can pass it using shiftout?

Like this: (shiftOut(dataPin, clockPin, MSBFIRST, B01001000); (which will illuminate LEDs on pin 2 and 5)


Here is my string building...

int latchPin = 8;
int clockPin = 12;
int dataPin = 11;

String byte_string;
int number_array[2] = {2,5}; //Shift register pin numbers to set to 1

void setup() {
  Serial.begin(9600); //For debugging
  pinMode(latchPin, OUTPUT);
  pinMode(clockPin, OUTPUT);
  pinMode(dataPin, OUTPUT);


void loop() {
    byte_string = "B";
    //Loop 8 times to capture 1/0 value for register pins
    for(int x = 1;x<=8;x++){
        byte_string = byte_string + "1";
        byte_string = byte_string + "0";
    //Display the dynamic string
  //Check if any of the array numbers exist in the 0 thru 8 loop
  bool checkExist(int iteration, int array[]){
    bool result;
    for(int j=0;j<=sizeof(array);j++){
      if(iteration == array[j]){
        result = true;
        result = false;
      return result;

if you have a var that indicates the arraySize of the array holding the numbers (this can vary over time) you could create a byte that holds the value to be shiftedOut().

something like

void doLeds()
  uint8_t leds = 0;
  for (int i = 0; i < arraySize; i++)
    leds |=  1 << array[i];
  shiftOut(dataPin, clockPin, MSBFIRST, leds);

I tried this method but it doesnt seem to work for my application. I have a String "B01001000" that is exactly what I want to shiftout. I just cant seem to get the same into byte form.


Thanks for the reply.

the code
leds |= 1 << array
uses a bitmask to set the bits. Normally bits are count [76543210] but I see in your code that you count them [12345678]. My mistake :wink:
try this
* *void doLeds() {   uint8_t leds = 0;   for (int i = 0; i < arraySize; i++)   {     leds |=  1 << array[i];   }   Serial.println(leds, BIN);  // for debug purpose   shiftOut(dataPin, clockPin, LSBFIRST, leds);  // changed MSBFIRST => LSBFIRST to change direction }* *
* *void doLeds() {   uint8_t leds = 0;   for (int i = 0; i < arraySize; i++)   {     leds |=  1 << (8-array[i]);  // change n => 8-n   }   Serial.println(leds, BIN);  // for debug purpose   shiftOut(dataPin, clockPin, MSBFIRST, leds); }* *

Wow....Thats amazing. Thanks for the help man. This worked perfectly for what I was trying to do. What exactly is the |= operation? (Forgive me. I am a vb and java programmer attemting C++)

Thanks again.


x |= y is a shorthand note for x = x | y

The | operator is a bitwise or
The << operator is a bitshift

similar C++ has
x += 3 => x = x + 3
y %= 5 => y = y % 5

z++ => z = z + 1
z-- => z = z -1

These are just shorthand notes for the programmer, but thanks to it the following is complete legal C

a = ++x + x++;

The use of bitwise operators are important when doing IO and setregisters etc. See also - Arduino Reference - Arduino Reference for doing lowlevel digitalWrite()

Thanks for the lesson and great links! You have been extremely helpful.

If you are trying to set & clear individual bits, you can do it like this also:

start with
byte x = B10101010; // 7 is usually treated as MSB, 0 as LSB, this equal 0xAA also
// set bit 7 Lo
x=x & B01111111; // only bit 7 is cleared by ANDing it with 0, rest retain their value
// x now eqauls B00101010
// set bit 2 Hi
x=x | B00000100; only bit 2 is changed by ORing with 1, rest retain their value
// x now equals B00101110

// now shift it out
shiftout (datapin, clockpin, MSBFIRST, x);