If statement doesn't work with array

[safa1]

I have the code below.

In setup i turn a integer value into binary array.

In loop:

And I want to blink led according to number my led doesn't turn off. only on it is.

Please help thank you.

//---------------var-----------------
char binary[9];
char* string;
//---------------var end-------------
void setup()
{
  Serial.begin(115200);
  for(byte i = 0; i < 7; i++){
    pinMode(i+13,OUTPUT);
  }

  byte someValue = 13; //For this example, lets convert the number 20
  
  
  binary[9] = {0}; //This is where the binary representation will be stored
  someValue += 128; //Adding 128 so that there will always be 8 digits in the string
  itoa(someValue,binary,2); //Conver someValue to a string using base 2 and save it in the array named binary
  string = binary + 1; //get rid of the most significant digit as you only want 7 bits


  Serial.println(string); //print out our string.
  for(byte i = 0; i < 7; i++){
    digitalWrite(i+13,string[i] - '0'); //write to the pin (the - '0' converts the bit of the string to HIGH or LOW)
  }
}

void loop()
{
for(byte i = 0; i < 7; i++){

Serial.println(string[i]);
  
if(string[i]==1){digitalWrite(D0,HIGH);}


else if(string[i]==0){digitalWrite(D0,LOW);}

delay(1000);


}

delay(5000);
Serial.println(string);

}

[paulpaulson]

Hello
Post your sketch in so called code tags "</>" to see how we can help.
Does the compiler likes the sketch?

[lastchancename]

I doubt a String can be compared with a numeric constant the way you’re thinking.

[safa1]

Sorry im new i fixed it

[paulpaulson]

Hello
My compiler doesn´t like the posted sketch.

[anon73444976]

@safa1

char binary[9];
...
...
  binary[9] = {0}; 

Oops.

[safa1]

could you explain my mistake im new
and what should i do to fix it

[anon73444976]

You have a nine element array, with elements numbered 0...8 inclusive.
There is no element 9.

Edit: what does "string" point to?

Edit:edit: found it

[anon73444976]

@lastchancename

I doubt a String can be compared with a numeric constant the way you’re thinking.

...but there are no Strings.

[Idahowalker]

once you've tended to the suggestions you can try to work with the following ideas to do the thing.

int x = 0b100101;
// First bit
(x >> 0) % 2; // 1
// Second bit
(x >> 1) % 2; // 0
// Third bit
(x >> 2) % 2; // 1

and so on and so forth

another way

int value = 26;
for (int i = 0; i < 8; i++) {
    if ((value >> i) & 0x01) {
        // Bit i is 1

    }
    else {
        // Bit i is 0

    }
}

[Whandall]

@safa1

  if(string[i]==1) {
     digitalWrite(D0,HIGH);
  } else if(string[i]==0) {
    digitalWrite(D0,LOW);
  }

This will not work, you have chars in the array '0' and '1', not the numbers 0 and 1.

The if of the else is not necessary, there are only two possibilities.

[lastchancename]

if(string[i]==1){

Close enough… a char certainly could contain a non-ASCII character, but there’s probably a better way to do it -

You’ll get there.