Escape sequences are producing unexpected output in C [duplicate]

Answer 1

Because you are writing everything on the same line after the first \n :

 printf( "\nab" ); // -> ab on a new line

Line status : ab

 printf( "\bsi" );

It means ab minus the b because of the backspace then you add si to the line.

Line status : asi

Finally :

 printf( "\rha" );

Means that you go back at the beginning of the line and write ha.

It leads to hai.

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\r is carriage-return, not linefeed.

Answer 2

First printf will print ab then \b will make backspace and print si so your final print becomes asi.

Now in last \r will move the cursor at beginning and prints ha which will replace first two character and makes final string as hai(ha from last statement and i from earlier result.)

Answer 3

\b is for backspace. That is, it removes the character just written.
\r is to return the cursor back to the left most position in the current line.

Let z represent cursor.
After statement one.

\n
abz

After backspace.

az

After statement two.

asiz

After return.

zbi

After statement three

haiz

Answer 4

\r is actually carriage return (which takes the cursor to the start of the line).

Your program outputs a new line (\n) followed by "ab" then backspace (\b) (over the b) "si", so you now have "asi" on the screen.
The \r takes the cursor to the start of the line and then outputs "ha" leaving "hai" on the screen.

Answer 5

The first instruction will print ab on a new line (\n) :

>ab

The second instruction will make a backspace (\b) before printing si :

>asi

Then the last one will make a carriage return (\r) before printing ha :

>hai

Answer 6

No, '\r' is not linefeed, '\n' is actually linefeed. It just happens to be so that many modern systems uses '\n' as both linefeed and carriage-return, making it "newline".

'\r' is carriage-return, which returns the cursor to the start of the line.