将int变量与字符串连接会导致奇怪的输出

Question

I have an arduino program in mind that i am trying to make.

Its purpose: Read digital pins from 2, to 11. Print the pin number, and "1" if pin is high, or "0" if pin is low.

This is what i've tried to do:

void loop() {
  for(int i = 2; i<12; i++){
    if(digitalRead(i) == HIGH){
      Serial.println(i + "1");
    }
    if(digitalRead(i) == LOW){
     Serial.println(i + "0");
   }
  }
}

The output should be "21" if pin 2 is HIGH , or "20" if pin 2 is LOW . The same applies to the other pins.

Instead , all it prints is

Ò>Tm_°

>Tm_°

>Tm_°

Tm_°







Ò>Tm_°

>Tm_°

>Tm_°

Tm_°

Any advice on how i can get this to work?

Answer 1

What happen with your code ?

Serial.println(2 + "1") won't give you 21 in C (in this case used for Arduino).

You are trying to concatenate an integer and a string directly and it is not valid in C (or almost programming language).

Solution:

void loop() {
  char pin_display;
  for(int i = 2; i<12; i++){
    if(digitalRead(i) == HIGH){
      pin_display = i + 0x30 //convert to Ascii
      Serial.print(pin_display);
      Serial.print("1");
    }
    ...

Answer 2

As mentioned in the other answer, the problem happens in Serial.println(i + "1") . This expression is evaluated as int + pointer that results in a corrupted pointer . A short way to fix that is creating a String from the integer variable: Serial.println(String(i)+"1") . This expression is evaluated as String + pointer that results in a valid String object.

Corrected code:

void loop() {
    for(int i = 2; i<12; i++){
       if(digitalRead(i) == HIGH){
          Serial.println(String(i) + "1");
       }
       if(digitalRead(i) == LOW){
         Serial.println(String(i) + "0");
       }
    }
}