scanf(), ......see the fun...whats the reason?

sateeshgalla · 08-22-2005, 06:08 AM

Hi,

I am Sateesh. I wrote a simple C code. it compiles well. but when run, it goes errotically.
please check out the code and give the reason behind it why its work like that and how to correct it.

the error is it asks for character only twice. and gives output as

$ a:a b: c:b
$

Here is the code.

int main()
{

char a,b,c;
scanf("%c", &a);
scanf("%c", &b);
scanf("%c", &c);
printf("a:%c\t b:%c\tc:%c\n",a,b,c);
return 0;
}

if add \n before %c , it works fine. why?

Thanks in advance.

bigearsbilly · 08-22-2005, 06:20 AM

It is working.
scanf is picking up the <RETURN> as a character.
it just picks up the character, YOU have to parse it!

It's often best to get a whole line and parse the input yourself.
Interpreting user input is always the trickiest part of any task

you'll notice if you type three characters *the* press return it works.

sateeshgalla · 08-22-2005, 06:37 AM

hi billy,
thak you very much. i understood.
when you scan a character, it takes any character \n or blank..everything is a character.
when you scan a integer, it doesnot take \n or blank character and when press enter also it doesnot accept \n and waits for integer value.