C: read struct fraction in procedure

Question

I'm trying to create a program which reads a number as type fraction using procedures. That's my code (the compiler gives me an error in readFraction() line 9):

#include <stdio.h>

typedef struct {
  int num,den;
} fraction;

void readFraction(fraction * f) {
  printf("Insert a fraction: ");
  scanf("%d/%d", &*f.num, &*f.den);
}

void writeFraction(fraction f) {
  printf("Fraction inserted: %d/%d\n", f.num, f.den);
}

void main() {
  fraction a;

  readFraction(&a);

  writeFraction(a);
}

This is the error message:

$ gcc fraction.c 
fraction.c: In function ‘readFraction’:
fraction.c:9:21: error: ‘f’ is a pointer; did you mean to use ‘->’?
    9 |   scanf("%d/%d", &*f.num, &*f.den);
      |                     ^
      |                     ->
fraction.c:9:30: error: ‘f’ is a pointer; did you mean to use ‘->’?
    9 |   scanf("%d/%d", &*f.num, &*f.den);
      |                              ^
      |                    

I know I could solve this using something like readFraction(&a.num,&a.den); or a=readFraction(); and the function returns the fraction but I'd like to know if I could just pass the fraction as a parameter.

Answer 1

The problem with your code is in the relative precedence of C operators. Structure-member selection has higher precedence than the dereferencing operator, so the sub-expression *f.num means "the object to which f.num points", not "member num of the object to which f points." The former is erroneous because f is a pointer, not a structure or union.

You could override the default precedence by adding parentheses:

  scanf("%d/%d", &(*f).num, &(*f).den);

, but it would be more idiomatic to use the indirect member selection operator:

  scanf("%d/%d", &f->num, &f->den);

Since we're talking about precedence here, it is worth noting that the -> operator has higher precedence than &, so no parentheses are required in that case.