My code doesn't work

[yekta_ino]

Hello, I was coding for a telegraph in web editor, but code doesnt work. The code is here:

int kont = 2;
int yazi[100];

void setup() {
  pinMode(kont, INPUT);
  Serial.begin(9600);
}

void loop() {
  for(int i = 0; i<101; i++){
    int ba=digitalRead(kont);
    if ba == 1{
      delay(200);
      if ba == 0{
        yazi[i] = 0;
        Serial.print(yazi[i]);
      }
      else{
        yazi[i] = 1;
        Serial.print(yazi[i]);
      }
    }
  }
  
}

And also the terminal:

/tmp/568693805/sketch_jun10c/sketch_jun10c.ino: In function 'void loop()':

/tmp/568693805/sketch_jun10c/sketch_jun10c.ino:12:8: error: expected '(' before 'ba'

if ba == 1{

^~

Error during build: exit status 1

I am sure with connections, but still doesnt working.

[anon57585045]

@yekta_ino

   if ba == 1{

C language requires brackets () around the boolean expression in an 'if' statement.

[TomGeorge]

Hi, @yekta_ino
Welcome to the forum.

What model Arduino are you using?

Can you please tell us your electronics, programming, arduino, hardware experience?

Thanks.. Tom..

[yekta_ino]

like this?

if ba = 1(){
   delay(9999);
}

[yekta_ino]

I am using uno, I am not proffesional in Arduino but I had a simple education

[madmark2150]

if (b == 1)
{...}

[yekta_ino]

thanks

[TomGeorge]

Hi,

If Statement (Conditional Statement) | Arduino Documentation

Have you programed an Arduino before?

What model Arduino are you using?

Thanks... Tom..

[anon56112670]

@yekta_ino

if ba == 1{
      delay(200);
      if ba == 0{

If it wasn't zero 200 milliseconds ago, its value is unlikely to have changed.

[MaximoEsfuerzo]

@yekta_ino

    if (ba == 1) {
      delay(200);
      if (ba == 0) {
...

If ba equals 1 it can never equal 0 at the same time then the second comparison will never be true.

[herbschwarz]

int yazi[100]; // reserves 100 addresses
...
for(int i=0; i<101; i++) // counts 101 numbers

[xfpd]

@anon57585045

brackets ()

I respect your knowledge on all subjects, @anon57585045, but may I sxy this:
() parentheses
[] brackets
{} braces

[camsysca]

{}[]()

The set of brackets, made up of subsets curly, square, and round.

[xfpd]

@camsysca

curly, square, and round

Ha. No.

"parenthetically" is not "round bracketly," et cetera, ibid.
Those terms are taught to those who can not remember or correctly pronounce words easily.
[edit] ... but this is not meant to point at you... it is pointed at whomever taught your teachers.

[camsysca]

Oh, I'm fully with you, should have winked as part of the comment. That came from my nephew's school. Sigh.