Wrong Output of a C program without errors

Hi guys i just wronte this small program in C using the Notepad++ and Cygwin. So the code is the following:

#include <stdio.h>

int main()
{
        int c, i, countLetters, countWords;
        int arr[30]开发者_C百科;

        countLetters = countWords = 0;
        for(i = 0; i < 30; ++i)
            arr[i] = 0;

        while(c = getchar() != EOF)
                if(c >= '0' && c <= '9')
                    ++arr[c - '0'];

                else if (c == ' ' || c == '\n' || c == '\t')
                    ++countWords;

                else
                    ++countLetters;

        printf("countWords = %d, countLetters = %d\n",
        countWords, countLetters );
}

but instead of counting words the program counts words as letters and print them as letters and words = 0... where am i wrong because even my teacher couldn`t give me an answer...

Try using curly brackets and the c = getchar() needs parentheses.

while((c = getchar()) != EOF) {
      ^             ^
     /* Stuff. */
}

The error is here:

while(c = getchar() != EOF)

You need to enclose the assignment in parentheses, like so:

while( (c = getchar()) != EOF)    /*** assign char to c and test if it's EOF **/

Otherwise, it is interpreted as:

while(c = (getchar() != EOF))    /** WRONG! ***/

i.e. c is 1 for each char read until the end of file.

The solution:

change while(c = getchar() != EOF), to while((c = getchar()) != EOF)

Reason:

!= has higher priority than =

Hence ,

getchar() != EOF

evaluates to be false and thus becoming

while(c=1) ==> while(0).

Hence, the loop gets iterated with c=1 ,what ever your input be. (except EOF).

In this case Your expression is always evaluates to be false.

since,

if(c >= '0' && c <= '9') is if(1>=48 && 1<=57) and it is always false.

Also,

else if (c == ' ' || c == '\n' || c == '\t')

will evaluated to be false.

Hence the else part countLetters++ will be executed for all inputs!

Resulting in the case as you prescribed.