Implementing fgetc; trying to read word by word

Question

I am trying to read word by word, and below is the logic that I have adopted. This is reading in the words fine, except when it gets to the last word in a line, in which it stores the last word of the current file AND the 1st word of the next new line. Could somebody tell me how I can get this to work?

int c;
int i =0;
char line[1000]
do{
    c = fgetc(fp);

    if( c != ' '){
        printf("%c", c);
    line[i++] = c;

    }else if((c == '\n')){

//this is where It should do nothing

    }else{
    line[i] = '\0';
    printf("\\0 reached\n");//meaning end of one word has been reached
    strcpy(wordArr[counter++].word, line);//copy that word that's in line[xxx] to the struct's .word Char array
    i=0;//reset the line's counter

    }//if loop end



} while(c != EOF);//do-while end

fp is a file pointer.

HI BABY TYPE MAYBE
TODAY HELLO CAR
HELLO ZEBRA LION DON
TYPE BABY

I am getting (w/o quotes)

"HI"
"BABY"
"TYPE" 
"MAYBE
TODAY"

Answer 1

This works for me (on Linux):

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

int
main(int argc, char **argv)
{
        char c;
        size_t i = 0;
        FILE *file = NULL;
        char buffer[BUFSIZ];
        int status = EXIT_SUCCESS;
        if (argc < 2) {
                fprintf(stderr, "%s <FILE>\n", argv[0]);
                goto error;
        }
        file = fopen(argv[1], "r");
        if (!file) {
                fprintf(stderr, "%s: %s: %s\n", argv[0], argv[1],
                                strerror(errno));
                goto error;
        }
        while (EOF != (c = fgetc(file))) {
                if (BUFSIZ == i) {
                        fprintf(stderr, "%s: D'oh! Write a program that "
                                        "doesn't use static buffers\n",
                                        argv[0]);
                        goto error;
                }
                if (' ' == c || '\n' == c) {
                        buffer[i++] = '\0';
                        fprintf(stdout, "%s\n", buffer);
                        i = 0;
                } else if ('\r' == c) {
                        /* ignore */
                } else {
                        buffer[i++] = c;
                }
        }
exit:
        if (file) {
                fclose(file);
        }
        return status;
error:
        status = EXIT_FAILURE;
        goto exit;
}

Answer 2

change if( c != ' ') to if( c != ' '&&c!='\n')

this should fix the problem

Answer 3

Try this;

 if((c == '\n') || (c == '\r'){ 

Answer 4

The encoding of the line terminating character is different from one operating system to another. In Linux, it is simply '\n', while in Windows and DOS it is '\r\n'. So, depending on your target OS, you may need to change your statement in something like:

if((c == '\r' || (c == '\n'))
{
   //...
}

EDIT: after looking closely, I think that what you're doing wrong is that the first if statement is true even when you read the \n, so you should handle it this way:

if((c != ' ') && (c != '\n')){
    printf("%c", c);
    line[i++] = c;
}
else if((c == '\n') || (c == '\r')){

//this is where It should do nothing

}
else{
   //...
}

Answer 5

Look at this:

if(c != ' ') {
    // ...
} else if(c == '\n') {
    // WILL NEVER BE REACHED
}

If c == '\n', then c != ' ' is also true, which means the second block will be skipped, and the first block will run for all '\n' characters, (i.e. they will be printed).

Other answers about line endings are wrong. C FILE *s not opened in binary mode will take care of EOL for you. If you have a file from DOS and you read it on Unix it might create problems, but I doubt that's your problem here, and if it was handling it could be a little more complicated than the answers here show. But you can cross that bridge when you reach it.