random number popups....

[i_can_do_this]

while i was bored and randomly typing code in and compiling it for no reason, this caught my attention:

#include <stdio.h>

int main()
{
    int c;
    
    while((c = getchar()) != EOF)
          printf("%d\n", (c = (c - '0') + 2));
}

shouldnt this equal the original number added by 2? well it does it show the result but it always has a -36 after it on the next line. i have no idea why this happens , i even tried it in different ways like:

c = c + 2;
printf("%d", c);

can somebody please explain this?

[ZuK]

    while((c = getchar()) != EOF)

Search the forum for an answer about why this is not a proper way to control an input loop.
HINT: you print EOF

Kurt

[Salem]

> Search the forum for an answer about why this is not a proper way to control an input loop.
Seems OK to me - you're thinking of feof()

> HINT: you print EOF
Mmm, I don't think so.

> but it always has a -36 after it
'\n' is 10
'0' is 48
2 is 2
Can you make -36 yet?

[ZuK]

Seems OK to me - you're thinking of feof()

Guess I did. Sorry.
Kurt

[ssharish2005]

avoid this by the following code

#include <stdio.h>

int main()
{
    int c;
    
    while((c = getchar()) != EOF)
    {
          printf("%d\n", (c = (c - '0') + 2));
          while((c = getchar())!= '\n' && c != EOF);
    }    
}
/*myouptut
a
51
s
69
d
54
f
56
g
57
h
58
j
60
k
61
*/

ssharish2005

[quzah]

Avoid what with that code? You write the oddest "fixes" I have ever seen.


Quzah.

[Salem]

> avoid this by the following code
Ah, you mean

while((c = getchar()) != EOF && c != '\n' )

[ssharish2005]

well, my intention here was to clear the input buffer when it enters the loop back again so that it wont agian read the '\n' char which was left in the input buffer. which actually avoids the printing the -36

and i am sorry i havn't included return value in the main. that was my bad

ssharish2005

[CornedBee]

Your bad was swallowing the entire input line after the first character, thus only printing the first line of each character.

[i_can_do_this]

i know if i dont use eof as a condition then it works and it doesnt count the '\n'. but how can i make the thing loop and not just terminate only after 1 input.

[quzah]

Well for starters, you could take the time to actually explain what it is you're trying to do. See, something like this:

"I want to read a line of text, and display each character's decimal, hex, and character representation on a seperate line. I want to repeat prompting for and display output until a blank line or EOF is entered."

Or whatever it is you actually are trying to do. I hate having to teach people how to think or ask a simple question. Go read your two posts. They are in no way related, and your second post isn't even a complete thought.


Quzah.

[i_can_do_this]

ok heres exactly what im trying to do. adding 2 to a number coming from input. i want this thing to loop so it will always add 2 to the input number and i wanna know how to stop the -36 from always showing up after the result.

[itsme86]

There are other, better ways to do it. What if the input doesn't even happen to be a number? Try this:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  char buf[BUFSIZ];
  int num;

  while(fgets(buf, sizeof(buf), stdin))
  {
    if(sscanf(buf, "%d", &num) == 1)
      printf("%d\n", num + 2);
    else
      fputs(buf, stdout);
  }

  return 0;
}

That will add 2 to any input number, but if the input isn't a number it'll just regurgitate what you typed.