What's wrong with my C character and line counting program?

Question

I was reading Kernighan Ritchie and there's this Character counting program, so I tried implementing

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

int main()
{
char c;
int i;
c = getchar();
while (c != EOF)
    i= i + 1;
printf("%d",i);
}`

When I compile and run this code, after I enter some characters there's no output after that. No number is printed and I can't figure out why. The code looks fine. I also tried using scanf() but the same thing happened.

The next example was for counting lines in input and the same problem was there too.

Answer 1

In your code, when you're hitting

 i= i + 1;

the initial value of i is indeterminate. So, the whole program invokes undefined behavior. You need to initialize i.

To elaborate, i being an automatic local variable, unless initialized explicitly, the content is indeterminate. Using an indeterminate value in this case will lead to UB.

That said, a char is not able to hold a value of EOF, change the type to int.

---

After that, you're wrong in the logic. getchar() is not a loop on it's own, you need to keep calling getchar() inside the while loop body to update the value of c, used in the condition check in while.

Answer 2

int main()

Good. Not main() and not void main(). Keep it this way.

{
char c;

Bad. getchar() returns an int.

int i;

Bad. Not initialised, value is indeterminate. You are going to increment -7445219, or perhaps a kidney pie, who knows.

c = getchar();

Ok you have read a single character.

while (c != EOF)

But you cannot compare it to EOF because EOF doesn't fit in a char.

    i= i + 1;

Looks like you forgot to do something in the body of the loop. Something that is going to change c perhaps so that your loop has a chance to finish.

printf("%d",i);

It is recommended to add a newline to the last line of your output.

}

That's all folks.

Answer 3

You only need to add an inicialization for "i"

int i = 0;

Answer 4

You missed the initialization:

int i = 0;