serial.println(String) prints "#" instead of the string - c

Question

i'm just trying to recieve a string using a serial and to send this string back. So when i send a string to an arduino over a serial the arduino should automaticly send this string back.
i created this code:
String test;
void setup(){
Serial.begin(9600);
Serial.setTimeout(2);
test = "null";
}
void loop(){
if(Serial.available()){
test = Serial.readString();
}
Serial.println(test);
}
I guess it is not that difficult to understand. However now the arduino will always print a "#" instead of the variable test. My connected serial device is a bluetooth modul. (hc-06)
What did i do wrong?
Thank you!
(i also ran this code in the arduino emulator 123D Circuits. There it worked just fine)

Answer 1

You need change your code. Move println into if statement.
Try increase timeout interval, 2ms is not enough, good value (at 9600) lies above 10ms. Theoretically timeout should be at least 3.5 characters long and for current speed this equals ~0,4 ms. But in practice higher values are used.
String test;
void setup(){
Serial.begin(9600);
Serial.setTimeout(10);// or more
test = "null";
}
void loop(){
if(Serial.available()){
test = Serial.readString();
Serial.println(test);// moved into if
}
}
Update: Another simple solution to return characters back looks like:
void loop(){
if(Serial.available()) Serial.write(Serial.read());
}
Update 2: Had similar issue with BLE module HM10 (clone, not official). It was sending about 15 dummy bytes prior to any array. And i didn't solve it. But if weired bytes always the same you can make a simple trick using String.remove():
if(Serial.available()){
test = Serial.readString();
test.remove(0,5);
// test.remove - add code to remove last character
Serial.println(test);
}
Also try another terminal.