Thread: Why doesn't this work? (simple)

> I would have thought the following code would print out
> integers if they were inputted......

> printf("%d\n", c);

All you're doing here is displaying the decimal value of the character that has been entered. If you're after numbers then yes, c-'0' will work.

For example, let's say that '0' has the decimal (ascii) value of 35. If I enter a '1', then it's value is 36, and '2' is 37 and so on.
That being the case:

37 - 35 = 2 (ie: '2' - '0' = 2)

Not a hard concept.

> variable 'c' should be a char and the format specifier will need
> to be a %c:

No. Because they want to display the numerical value. As such, %d is what they want.

Quzah.

Thanks quzah, that makes sense.

Though i'm still puzzled by the output of this code:

Code:

#include <stdio.h>

int main()
{

int c;

while ((c = fgetc(stdin)) != 'Q')
  printf("%d\n", c - '0');

return 0;

}

I would have thought when running it, if i entered say 5, it would just print 5 right back out........ instead it prints 5 then underneath the 5 it prints -38.......

where is the -38 coming from?