Escape sequences are producing unexpected output in C

Question

I am a beginner in C programming language, recently I have studied about escape sequence.

\n means newline
\b means backspace
\r means carriage-return

When I am applying these on the following program, then I am getting output as hai, Can anyone please explain, how?

main()
{
    printf("\nab");
    printf("\bsi");
    printf("\rha");
}

Your assumption may not be true, and the output may varies, see [Usage of \b and \r in C](http://stackoverflow.com/q/17236242/1009479) — Yu Hao, Aug 07 '13 at 08:48

score 13 · Accepted Answer · edited Aug 07 '13 at 08:57

13

\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.

edited Aug 07 '13 at 08:57

Werner Henze

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answered Aug 07 '13 at 08:46

noz

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score 5 · Answer 2 · answered Aug 07 '13 at 08:47

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

answered Aug 07 '13 at 08:47

zakinster

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score 4 · Answer 3 · answered Aug 07 '13 at 08:46

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.)

Pierre Fourgeaud · Answer 4 · 2013-08-07T08:59:00.030

3

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.

\r is carriage-return, not linefeed.

edited Aug 07 '13 at 08:59

answered Aug 07 '13 at 08:48

Pierre Fourgeaud

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score 2 · Answer 5 · answered Aug 07 '13 at 08:46

2

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.

answered Aug 07 '13 at 08:46

Some programmer dude

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score 1 · Answer 6 · answered Aug 07 '13 at 08:49

\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

score 0 · Answer 7 · answered Aug 07 '13 at 09:18

Here in your code

\r

\r means Carriage Return, which means: return the cursor to the beginning of the line in more simple words we can say that it's deleting each character from the active position upto the beginning.

\n
```
\n mean New Line
```
\b
```
\b mean Backspace
```

Let's take it one step at a time:

<new line>ab<backspace>si<carriage return>ha

First, handle the backspace. Note that even though it is "non-erase", the next character to be output would overwrite what was backspaced over:

<new line>asi<carriage return>ha

Now, a carriage return means to go back to the beginning of the line. So the "ha" overwrites the "as" in "asi:

<new line>hai

Now, the cursor is currently sitting on the i, so the next character to be output would overwrite i.

Escape sequences are producing unexpected output in C

7 Answers7

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