Counter depending on getchar return value gives wrong oputput

Question

#include <stdio.h>

int main(void)
{

    char ch;
    int count;

    while((ch=getchar())!=EOF){
        if(ch==' '){
           count++;
        }
    }
    printf("total words is %d",count);
    return 0;
}

my C code is here ,I am confused where I put ctrl+Z directly in windows ,it turn out count is 2?why it go loop for two and where comes the space?

Assigning what is returned from `getchar()` to variable having type `char` is bad because it will truncate the data and it make distinguishing of a charcter and `EOF` impossible in typical environment. Using `int` is good. — MikeCAT, Jun 12 '16 at 13:39
Related, if not a dupe to: http://stackoverflow.com/q/35356322/694576 — alk, Jun 12 '16 at 13:42
Please read manuals for all library functions that you want to use. — Kerrek SB, Jun 12 '16 at 13:42

Sourav Ghosh · Answer 1 · 2016-06-12T13:55:17.950

The biggest issue here is, you've left count uninitialized. So, whether the if is success or not, you'll end up reading a unitialized local variable which invokes undefined behavior.

You should always initialize the local variables, like

 int count = 0;

That said, getchar() returns an int. Storing that into a char is wrong because in case the function returns EOF (as you expect here), it'll not fit into a char.

Regarding EOF marco, quoting C11, chapter §7.21.1

EOF
which expands to an integer constant expression, with type int and a negative value, that is returned by several functions to indicate end-of-file, that is, no more input from a stream;

So, you should change the type of ch from char to int, like

int ch = 0;

instead of saying thank u, you can accept an answer if it solves your problem. — mssirvi, Jun 12 '16 at 13:47

Counter depending on getchar return value gives wrong oputput

1 Answers1