How to correctly assign a pointer to a structure variable in this case?

Question

Can someone please point out what's the error in this assignment of pointer to a structure?

Brief background- I have a structure called fraction, I have two variables of type fraction (f1 &f2). I also have an array of structures called num. It all worked well so far until I tried to create a pointer of type struct fraction and assign it f1's address.

#include <stdio.h>

struct fraction
{
    int nume;
    int deno;
};

struct fraction *fp;
struct fraction f1,f2;
struct fraction num[3];
fp=&f1;

int main()
{
    f1.nume=22;
    f1.deno=7;
    int i,x=5,y=6;
    f2=f1;

    printf("\n Fraction 1 numerator is: %d\n", f1.nume);
    printf("\n Fraction 1 denominator is: %d\n", f1.deno);

    printf("\n Fraction 2 numerator is: %d\n", f2.nume);
    printf("\n Fraction 2 denominator is: %d\n", f2.deno);

    //initialize array of structs
    for(i=0;i<3;i++)
    {
        num[i].nume=x;
        num[i].deno=y;
        ++x;
        ++y;
    }

    //print out array of structs
    for(i=0;i<3;i++)
    {
        printf("\n Numerator of fraction num %d : %d", i, num[i].nume);
        printf("\n Denominator of fraction num %d : %d\n", i, num[i].deno);

    }

    //try to use pointer to access field
    printf("\n Numerator of fraction f1 : %d\n", fp->nume);
    printf("\n");
}

I was thinking pointers could be assigned to variables of a structure type following the normal rules of pointer assignment with int, float etc. For instance- you could do:

int x=10;
int *y;
y=&x;

This points the pointer y to the address of variable x.

Then, following the same logic, why does the code above not work? I tried to verify my method online, and from this eskimo.com article, it says pretty much the same thing. Here's the error I get:

struct1.c:21:1: warning: type specifier missing, defaults to 'int'
      [-Wimplicit-int]
fp=&f1;
^~
struct1.c:21:1: error: redefinition of 'fp' with a different type: 'int' vs
      'struct fraction *'
struct1.c:17:18: note: previous definition is here
struct fraction *fp;
             ^

Many thanks for the help!

Solution 1

struct fraction *fp;
struct fraction f1,f2;
fp=&f1;

is indeed similar to

int x=10;
int *y;
y=&x;

However, did you try to use the latter y=&x; outside any functions like what you did with fp=&f1;? That would not compile because for both examples it is not legal for the assignment to be put there.

The solution is simple: put fp=&f1; inside a function, like main.

Solution 2

As @Yu Hao suggested either you can put fp=&f1; inside a function, like main Or

You can change your code to :

struct fraction f1,f2;
struct fraction *fp = & f1;
struct fraction num[3];