white space characters and scanf

**bthomson900** · 11-10-2010

I am studying for an exam at the moment (my first one for eng computing) and I came across this in my lecture notes:

'Care must be taken when reading characters as such with
scanf( ) as the 'white space' characters, viz. blank, tab (\t)
and newline (\n) are valid characters for the %c conversion.'

Correct me if I'm wrong but I understand that 'white space' is any multiple of blank, tab or new line spaces (so anything that is not occupied by characters basically i.e pressing spacebar, tab or enter whilst entering the input character stream).

What confuses me is how could blank, tab and newline be "valid characters" for conversion? What exactly does this mean? Could someone please provide an example to explain this?

I mucked around a bit and if I had two, character type variables declared and I entered "\n 5" into the input character stream, then the two variables are assigned \ and n respectively. I think I am interpreting this information the wrong way.

**laserlight** · 11-10-2010

Originally Posted by bthomson900

Correct me if I'm wrong but I understand that 'white space' is any multiple of blank, tab or new line spaces (so anything that is not occupied by characters basically i.e pressing spacebar, tab or enter whilst entering the input character stream).

Yes, so a white space character is one such character.

Originally Posted by bthomson900

What confuses me is how could blank, tab and newline be "valid characters" for conversion? What exactly does this mean? Could someone please provide an example to explain this?

The idea is that for most conversion specifiers, scanf will skip leading whitespace, but %c will not skip them, but rather match the character.

Originally Posted by bthomson900

I mucked around a bit and if I had two, character type variables declared and I entered "\n 5" into the input character stream, then the two variables are assigned \ and n respectively. I think I am interpreting this information the wrong way.

As you noted, whitespace characters can be entered by "pressing spacebar, tab or enter", so entering "\n" is wrong.

**bthomson900** · 11-10-2010

The idea is that for most conversion specifiers, scanf will skip leading whitespace, but %c will not skip them, but rather match the character.

Alright, I understand "skip leading whitespace", but what exactly do you mean by "%c will not skip them, but rather match the character". Could you provide a small example maybe?

Thanks

**laserlight** · 11-10-2010

Originally Posted by bthomson900

Alright, I understand "skip leading whitespace", but what exactly do you mean by "%c will not skip them, but rather match the character". Could you provide a small example maybe?

Here is a simple example:

Code:

#include <stdio.h>

int main(void)
{
    char c;
    scanf("%c", &c);
    printf("You entered '%c'\n", c);
    return 0;
}

Eneter whatever input you want, with say, a space in front. Observe the output.

**bthomson900** · 11-10-2010

Thanks mate. I think I've got it now.

I also added a few more variables of character type to the scanf function.

What I noticed was that any leading 'white space' (of any size) became was assigned to the first variable read. Furthermore, any subsequent white space between other characters entered into the input stream does not affect the assignment of any following characters.

i.e.

if I had three variables and entered: " 1 2"

the output generated would be: "You entered ' ','1' and '2'"

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