Question

In code walk-through into some others code I found *
What is this ?

here is the same prototype of that code and it’s not giving any error.
Only warning comes with gcc 4.5.2 But on windows it’s not giving warning and compiles properly. I can’t provide actual code due to IP of the company but I wanted to know this how and why this is correct?

#include<stdio.h>

typedef enum e
{
        a,
        b,
        c
}x,*y;

int main()
{
        x t=4;
        printf("x : %d\n",t);
        y m=5;
        printf("*y : %d\n",m);
        return 0;
}

Answer 1

It is equivalent to :

typedef enum e *y;

So y is an alias of enum e *, ie y m declares m as pointer to enum e. The following code works indeed as expected:

#include <stdio.h>

x t = 4;
printf("x : %d\n", t);
y m = &t;
printf("*y : %d\n", *m);