CODEWITHSUNDEEP
c
user1607425
user1607425

Reputation:

Reading chars in C

I wrote the following program to understand the behavior of EOF:

#include<stdio.h>

int main ()

    char c = 0;

    printf("Enter some chars, EOF stops printing:");

    while ((c == getc(stdin)) != EOF) 
        printf("c = %c, (int) c = %d\n", c, (int) c);

    return 0;
}

However, if I input something such as abcd I get this output:

c = a, (int) c = 97
c = a, (int) c = 97
c = a, (int) c = 97

Upvotes: 1

Views: 118

Answers (2)

unwind
unwind

Reputation: 399833

You must read the documentation better; getc() returns int, because EOF doesn't fit in a char.

Also, you're using both scanf() and getc(), which will make things confusing due to the input stream buffering.

Try something like this:

#include <stdio.h>

int main()

    int c = 0;

    printf("Enter some chars, EOF stops printing:");
    while ((c = getc(stdin)) != EOF) {
        printf("c = %c, (int) c = %d\n", c, c);
    }
    return 0;
}

I also added the missing } in your code, and removed the cast of c in the call to printf(), that's not needed now that c is int. This is the proper type for the %c formatting specifier too, by the way.

Upvotes: 1

cnicutar
cnicutar

Reputation: 182639

You've got a == instead of a = so you never store whatever getc returns:

while ((c == getc(stdin)) != EOF) {
          ^^

And of course c should be int, not char.

Upvotes: 7

Related Questions