Serial.print and char[] & null

Question

I don't write a lot of C, so I'm always discovering things I only partly understand in the Arduino variety. One of these is how Serial.print handles char[].

Does Serial.print always require a null terminator? In the case of (say) Serial.print(88, DEC); is a null implied?

What happens in the following cases?

char buff[3] = "ABC"; // warned because no room for a null
Serial.print(buff);   // will absence of a null cause overrun?

char buff[4] = "ABC"; // now there's a null
Serial.print(buff);   // will the null character be sent too?

Any references appreciated.

Answer 1

When passing a char* you need to also pass some way of telling the function where it ends. In C that is usually the null terminator and that's all a null terminator is for. If that is missing then you get a buffer overrun.

The null terminator is not sent.

Answer 2

The "Serial" is a object/class and that makes it possible to have more than one function for Serial.print(). According to the type of the parameter the proper Serial.print() is selected.

This is the include file for the Serial.print(): Print.h
And these are the possibilities:

• size_t print(const __FlashStringHelper *);
• size_t print(const String &);
• size_t print(const char[]);
• size_t print(char);
• size_t print(unsigned char, int = DEC);
• size_t print(int, int = DEC);
• size_t print(unsigned int, int = DEC);
• size_t print(long, int = DEC);
• size_t print(unsigned long, int = DEC);
• size_t print(double, int = 2);
• size_t print(const Printable&);

The reference for Serial.print() is short, but it explains pretty well that those different parameters are possible.

Your Serial.print(88, DEC); does not use a string of text, it uses the integer number 88.

As soon as a string of text is used, it should be zero-terminated.
The line char buff[3] = "ABC"; is indeed wrong, as you already wrote. That is one of the things of the 'c' language. The "ABC" gets a zero-terminater and that makes 4 bytes.