What is wrong with this code of file handling in c?

Question

I am trying to write content in a file from terminal. File is creating but content is not written into the file.

    #include<stdio.h>
    #include<stdlib.h>
    #include<math.h>
    int main(int argc, char *argv[])
    {
        FILE *fp;
        fp=fopen(argv[1],"w");
        char ch;
        while((ch=getchar())!=EOF)
        {
           putc(ch,fp);
        }
        fclose(fp);
        return 0;
    }

Sidenote: Always check return values of I/O functions, including `fopen` and `putc`. — user694733, Sep 19 '17 at 09:50
what happen if you display the character being read just before calling `putc()`? — JFMR, Sep 19 '17 at 09:51
Problem is not reproduce able. This code is working as expected for me. — svtag, Sep 19 '17 at 09:56
`getchar()` returns a `int`, not a `char`, so `char ch;` -> `int ch;`. — Jabberwocky, Sep 19 '17 at 09:59
You should check if `fopen(argv[1],"w")` returns `NULL`, that means the file could not be created for some reason. — Jabberwocky, Sep 19 '17 at 10:01

score 5 · Accepted Answer · answered Sep 19 '17 at 09:53

If you don't signal EOF (Ctrl+Z in Windows and Ctrl+D in Linux), then the loop will continue to execute until it receives that signal.

If you attempt to read the file with your own eyes while the program is still on execution, then the file stream will not have close (fclose(fp); will not have execute), thus the file will appear to you empty, even though the content will be shown to you, when the file stream closes.

score 1 · Answer 2 · answered Sep 19 '17 at 10:35

The following works fine:

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main(int argc, char *argv[])
{
    FILE *fp;
    fp=fopen(argv[1],"w");
    char ch;
    while(1)
    {
        ch = (char)getchar();
        putc(ch,fp);
        if(ch == '.') break;
    }
    fclose(fp);
    return 0;
}

What is wrong with this code of file handling in c?

2 Answers2