Relation of stdin and EOF in C

Question

A C quiz had these four questions that deal with fprintf and EOF. They weren't really explained in the section of the course related to these questions and I can't find good answers for them online.

Is the reason for the numbers printed on the screen and/or compilation fail the result of fprintf or its relation to EOF?

#include <stdio.h> 

int main(void) { 
    int i; 
    i = fprintf(stdin,"Hello!"); 
    printf("%d",i == EOF); 
    return 0; 
} 

Answer: the program outputs 1

#include <stdio.h> 

int main(void) { 
    int i; 
    i = fprintf(stderr,"Hello!"); 
    printf("%d",i == EOF); 
    return 0; 
}

Answer: the program outputs 0 to the stdout stream

#include <stdio.h> 

int main(void) { 
    FILE *f; 
    int i = fprintf(f,"Hello!"); 
    printf("%d",i == EOF); 
    return 0; 
}

Answer: the compilation or execution fails

#include <stdio.h> 

int main(void) { 
    FILE *f = fopen("file","w"); 
    int i = fprintf(f,"Hello!"); 
    printf("%d",i != EOF); 
    return 0; 
}

Answer: the program outputs 1

Answer 1

EOF is a macro that is very commonly equal to -1.

1. Q1 :
   • stdin is an input stream. Therefore using fprintf on it created a conflict, which apparently was handled by fprintf() and returned -1. ( as mentioned in the comments the return values in case of an error can be any negative number ).
   • In this sense the condition (i == EOF) returned true which is 1. which explains the answer.
2. Q2 :
   • stderr is an output stream, meaning printing into it was a success
   • i = ..., i contains the number of characters written. >0
   • i == EOF evaluates to 0.
   • 0 got printed
3. Q3 :
   • fprintf tries to print into a stream with the address f
   • f is a pointer with a garbage value, which means it might be pointing to a memory owned by the OS.
   • A write to that address resulted in a compilation or execution fails
4. Q4 :
   • fprintf does its job and write 5 chars into the file f
   • i gets initialized with the value 5
   • i != EOF evaluates to 5 != -1, which is true, the is stored as the number 1 in memory
   • printed value is 1