sprintf() in Arduino code

sprintf.ino

unsigned int i = 0;

void setup() {
  Serial.begin(9600);
}

void loop() {
  char buffer[50];
  sprintf(buffer, "the current value is %d", i++);
  Serial.println(buffer);
}

@RagBillySandstone you can't use Serial.println() in this way.

@RagBillySandstone You are not bypassing the buffer that way. In your code, Serial.println() had to return a buffer. This is not the case. The buffer is created to store the character of the string and keep them, until they are passed to println as a parameter. So your code doesn't make sense at all, since you've changed the whole logic with your change :)

hola como puedo en arduino
ingresando numeros por teclado que cree una cadena numerica
enves de que se sumen?