Serial.println printing just after the previous string in same line

Question

I wrote serial.println in my code but the string was printed in same line on serial moniter.

#include <WiFi.h>//for connecting esp32 to a wifi
#include <TinyGPS++.h>//to obtain gps data from neo-6m gps module 
#define WIFI_SSID "Airtel_pawa_4182"//wifi name in simple words
#define WIFI_PWD "Ahuja6230"//wifi password
TinyGPSPlus GPS;//creating a gps object from tinygps++ library
void setup() {
 Serial.begin(9600);
 //wifi setup{
 WiFi.begin(WIFI_SSID, WIFI_PWD);
 Serial.println("connecting to wifi");
 while(WiFi.status() != WL_CONNECTED){
 Serial.println(".");
 delay(1000); 
 }
 Serial.println("Successfully connected to ");
 Serial.print(WIFI_SSID);
 //}
 //gps setup{
 Serial2.begin(9600);
 delay(1000);
 Serial.println("ESP32-GPS Tracker");
 Serial.println("Initializing...");
 //}
}
void loop() {
}

In this code the string "esp32-gps tracker" is printing just after the previous string in same line on serial moniter.

I tried running code again and again but it didn't work...

I am using esp32 board(just a detail)

What do i do?

• Edgar Bonet: You need to be way more specific for your question to be answerable. Same line? Same as what? Please, post a minimal testable sketch showing the issue, tell us what you expected as an output, and tell us what the actual output was.
• Yug Ahuja: @EdgarBonet done

Answer 1

The previous print statements are these:

 Serial.println("Successfully connected to ");
 Serial.print(WIFI_SSID);

The SSID won't be terminated by a newline. Perhaps you meant:

 Serial.print("Successfully connected to ");
 Serial.println(WIFI_SSID);

?

Answer 2

To follow on from previous answer, your solution is to either:

• change the previous Serial.print() to a Serial.println() so that that text is followed by a line end, or
• add a \n to the front of the line you are having the problem with, so it starts with a new line, to make up for the "missing" new line in the previous Serial.print().

• Edgar Bonet: For consistency with what Serial.println() does, you should add "\r\n", not only '\n'. If you ever use a terminal emulator instead of the Arduino serial monitor, a bare '\n' will likely move the cursor one line down while staying on the same column.
• Nick Gammon: Depends on your operating system. In Linux a newline advances to the start of a new line. On Windows YMMV. It's very common in Linux code to use \n for a newline when outputting and not \r\n, which may in fact put some weird character (the carriage-return character) in your terminal window.
• Edgar Bonet: @NickGammon: On a Linux terminal, a \n alone is a vertical-only cursor movement. When a program outputs \n, the tty driver converts it to \r\n before it reaches the terminal. This should become clear if you try to use a terminal with this output post-processing disabled (stty -opost). A serial communication program like picocom, by default, disables all tty driver processing: the bytes from the Arduino reach the terminal unaltered. This plays well with Serial.println(), as it ends the line with \r\n.

Answer 3

When sending text through a serial line, it is customary to terminate each line with some sort of end-of-line marker: you print some text, and immediately after it the EOL to signify that "this is now a complete line". Accordingly, the Serial.println() method prints the requested text, immediately followed by the EOL marker (in this case, CR+LF).

From your question, you seem to be thinking that the EOL immediately precedes the text. This is not the case.

See JRobert's answer for the correct way to build a line from multiple print statements: println() should be use on the last piece, not on the first one.

• Yug Ahuja: Yes. Nice explanation. Now I know where I mistook. I thought that by using "print.ln()", the string will be printed in new line(I got this info from google). But it isn't like that print.ln is for terminating a line and starting a new line
• Yug Ahuja: Jrobert's comment was the first one to mark my mistake so I declared it as answer. But I also thank you for the explanation
• Nick Gammon: @YugAhuja println not print.ln you need to be very specific when coding.