Question

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fgetc does not identify EOF
fgetc, checking EOF

I have created a file and named it “file.txt” in Unix. I tried to read the file content from my C program. I am not able to receive the EOF character. Unix doesn’t store EOF character on file creation? If so what is the alternative way to read the EOF from a Unix created file using C.

Here’s the code sample

int main(){
File *fp;
int nl,c;
nl =0;
fp = fopen("file.txt", "r");
while((c = fgetc(fp)) != EOF){
  if (c=='\n')
    nl++;
}
return 0; 
}

If I explicitly give CTRL + D the EOF is detected even when I use char c.

Answer 1

This can happen if the type of c is char (and char is unsigned in your compiler, you can check this by examining the value of CHAR_MIN in ) and not int.

The value of EOF is negative according to the C standard.

So, implicitly casting EOF to unsigned char will lose the true value of EOF and the comparison will always fail.

UPDATE: There’s a bigger problem that has to be addressed first. In the expression c = fgetc(fp) != EOF, fgetc(fp) != EOF is evaluated first (to 0 or 1) and then the value is assigned to c. If there’s at least one character in the file, fgetc(fp) != EOF will evaluate to 0 and the body of the while loop will never execute. You need to add parentheses, like so: (c = fgetc(fp)) != EOF.