How to display characters read from a file using getc()

Question

When I try to read input from a file named "file1", my program correctly displays the number of characters in the file, but in an unrecognized character format. Below is the code

#include <stdio.h>
#include <stdlib.h>
void db_sp(FILE*);

int main(int argc,char *argv[])
{   
    FILE *ifp,*ofp;

    if(argc!=2) {
      fprintf(stderr,"Program execution form: %s infile\n",argv[0]);
      exit(1);
    }
    ifp=fopen(argv[1],"r");
    if (ifp==NULL) printf("sdaf");
    //ofp=fopen(argv[2],"w+") ; 
    db_sp(ifp);
    fclose(ifp);
    //fclose(ofp);
    return 0;
}

void db_sp(FILE *ifp)
{     
    char c;
    while(c=getc(ifp) !=EOF) {
      //printf("%c",c);
      putc(c,stdout);
      if(c=='\n' || c=='\t' || c==' ')
        printf("%c",c);
    }
}

Can you show what your file contains? What does your prinft print? — Yashas, Mar 22 '17 at 13:41

001 · Accepted Answer · 2017-03-22T13:53:59.333

2

The problem is here:

while(c=getc(ifp) !=EOF){

Because of operator precendence, this getc(ifp) !=EOF gets executed first. Then c = <result of comparison> gets executed. Which is not the order you want.

Use parentheses to force the correct order.

while((c=getc(ifp)) !=EOF) {

Other notes: getc returns an int so you should change the type of c to int. Also, if you fail to open the file, you still continue execution. You should gracefully exit on failure.

edited Mar 22 '17 at 13:53

answered Mar 22 '17 at 13:48

001

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And change `char c;` to `int c;` which is the type returned by `getc()` and required by `putc()`. – Weather Vane Mar 22 '17 at 13:51
@WeatherVane True. Updated. – 001 Mar 22 '17 at 13:54
@AjeyaSiddhartha for what it's worth `'\n'` is also of type `int`. – Weather Vane Mar 22 '17 at 13:56
The reason for getc to return an int and not a char, is that `'\xff'` is a valid character and shall not be misinterpreted as an eond of file. – Serge Ballesta Mar 22 '17 at 14:08

How to display characters read from a file using getc()

1 Answers1