'\n' characters before EOF

Question

char ch;
int n = 0;
FILE* fp;

fp = fopen("test.txt", "r");
while(!feof(fp)){
    n++;
    fscanf(fp, "%c", &ch);
    fprintf("%c", ch);
}
printf("%d\n", n);

test.txt below

abcd

I tried to count how many time does this while loop go by printing out integer n. I thought the result would be 4 cuz fp only contains 4 characters, "abcd". But actual result was 6. There are two newline character which is ascii code 10 before EOF.

I double-checked test.txt file and it contains no other character then abcd. Where did these 2 newline characters come from? I used vim editor.

You are using `feof` incorrectly: http://stackoverflow.com/questions/5431941/while-feof-file-is-always-wrong — William Pursell, Apr 03 '15 at 13:02

score 2 · Accepted Answer · edited May 23 '17 at 11:43

2

First, don't use scanf when getc will do. Second, stop using feof as if it is magic. (Why is “while ( !feof (file) )” always wrong?). Third, "abcd\n" is 5 characters, and your misuse of feof is causing you to reuse the final newline and overcount by one.

edited May 23 '17 at 11:43

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answered Apr 03 '15 at 13:04

William Pursell

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i didn't even press enter after that d and still it is abcd\n? – Castle_Dust Apr 03 '15 at 13:08
@Castle_Dust Some editors will insert the newline for you. Try `wc -c test.txt` to get a count, and `xxd test.txt` to see the content. – William Pursell Apr 03 '15 at 15:52

'\n' characters before EOF

1 Answers1