Treatment of \b in template strings to printf in C - c

Question

Can someone please explain or point me to a reference that explains why in the code snippets below the line printed in the first includes the whole word hello, while in the second it consists only of the letters he? I thought the backspace escape characters would delete the last three letters regardless of the newline escape. I've read the GNU documentation on printf but couldn't find anything on point.
int main(void)
{
printf("hello\b\b\b\n");
return 0;
}
int main(void)
{
printf("hello\b\b\b");
return 0;
}

Answer 1

This isn't really a question about C or printf. In each case the code does exactly what it says: in the first example it outputs the 9 characters h e l l o \b \b \b \n and in the second it likewise outputs 8 characters. If you write standard output to a file that's exactly what the file will contain. But if you write to a terminal, your terminal may handle \b by backspacing, and so your question is really about your terminal. You have not said what terminal you are using.
One common way for the backspace character to behave is by moving the cursor one position left, but not actually erasing the character in that position. The next character output will be written over it.
In your first example, you move the cursor back to the first l, but then you never write anything over it or any other character. Then \n doesn't change any character on the screen, but just moves the cursor to the next line. So you still see hello on the preceding line.
In your second example, the cursor is left over the first l when your program terminates. So the next program to write output to the terminal will write over that character (unless the next program's first write is \n or something like that). That next program might be your shell, and so the characters from your shell prompt probably overwrite the llo, so you don't see them.

Answer 2

Two key things to keep in mind:
Output to the console isn't immediately rendered when you use printf(): the output is buffered. Two ways to immediately (synchronously) flush the output buffer to the console is to print a newline via printf("\n"), or manually flush the buffer via fflush(stdout).
If you use the \b escape sequence, you will move the cursor one "space" back/left, but printing a newline character won't necessarily clear the rest of the line. If you were to manually print new characters over the old ones (i.e. printf("Hello\b\b\b \n")) the output would be more like what you're expecting.
Additionally, if you want to do some advanced command-line menus, graphics, etc, you'll need to use something like libncurses.
Finally, the following example should help give a better idea on the typically behavior involving your question.
#include <stdio.h>
#include <unistd.h>
#define DELAY (2) /* seconds */
int main(void)
{
printf("Example DELAY:\n");
printf("------------------------------------------------------------------------\n");
printf("hello\b\b\b\n");
sleep(DELAY);
printf("\n\n");
printf("Example 2:\n");
printf("------------------------------------------------------------------------\n");
printf("hello\b\b\b");
sleep(DELAY);
printf("...");
sleep(DELAY);
fflush(stdout);
sleep(DELAY);
printf("\n\n");
printf("Example 3:\n");
printf("------------------------------------------------------------------------\n");
fflush(stdout);
printf("hello\b\b\b");
fflush(stdout);
sleep(DELAY);
printf("...");
sleep(DELAY);
fflush(stdout);
sleep(DELAY);
printf("\n\n");
return 0;
}