Question

In the below example of pointer to structure

   #include<stdio.h>
   #include<stdlib.h>

   struct siva {
   int a;
   };

   typedef struct siva *s1;

    void main()
   {
    s1 b=(s1)malloc(sizeof(s1*));
    b->a=8;
    printf("\n The vlue is s1->a is %d",b->a);
    free(b);
  }

In the above code the variable b is not declared, but how this code works

In this line

   s1 b=(s1)malloc(sizeof(s1*));

How the memory is allocated for b and why s1* is used in sizeof().
what is the difference between s1 and s1

Answer 1

The following is a declaration of s1‘s type as a pointer to struct siva

typedef struct siva *s1;

The following is a pointer to s1 (e.g. pointer to pointer to struct siva)

s1*

The size given in malloc is wrong, it should be

s1 b=(s1)malloc(sizeof(struct siva));

As s1 points to struct siva. It works only because the address of the first element in the struct is the address of the struct, but this is basically a dangerous thing to do.

b is declared, in the same line it is assigned to:

    s1 b=(s1)malloc(sizeof(s1*));
//  ^^^^

Note that in C you are not allowed to initialize a variable from a function’s return value.