Thread: Accessing structure member through pointer using dynamic memory allocation

Hello,

Code:

#include<stdlib.h>
struct name {
   int a;
   float b;
   char c[30];
};
int main(){
   struct name *ptr;
   int i,n;
   printf("Enter n: ");
   scanf("%d",&n);
   ptr=(struct name*)malloc(n*sizeof(struct name));
/* Above statement allocates the memory for n structures with pointer ptr pointing to base address */
   for(i=0;i<n;++i){
       printf("Enter string, integer and floating number  respectively:\n");
       scanf("%s%d%f",&(ptr+i)->c,&(ptr+i)->a,&(ptr+i)->b);
   }
   printf("Displaying Infromation:\n");
   for(i=0;i<n;++i)
       printf("%s\t%d\t%.2f\n",(ptr+i)->c,(ptr+i)->a,(ptr+i)->b);
   return 0;
}

In the above code i would like to use (.) Operator in scanf and printf statements but i am getting some errors, please help me how can i use (.) operator in scanf and printf statement.


scanf("%s %d %f", &(*ptr+i).c, &(*ptr+i).a, &(*ptr+i).b);


printf ("%s %d %f\n", (*ptr+i).c, (*ptr+i).a, (*ptr+i).b);

Error:

[Error] invalid operands to binary + (have 'struct name' and 'int')

Use (*(ptr+i)). ... . You need to add the index to the pointer before dereferencing it. You may not need the outer parenthesis, I didn't try to compile the code.

You should see struct name *ptr as a variable that holds the address of another name struct. *ptr holds no actual data about the contains of a name struct itself only the location where the data should be stored. malloc was given the task to find a nice place somewhere where all that data could be physically stored and all *ptr has to do is remember the location of this data block.

Because *ptr itself has no members like a, b or c, *ptr only holds the location where they are stored, you cannot say ptr.a, ptr.b or ptr.c because they simply don't exist.

(ptr + i)->c could be read as: point to start location of data block + size of struct name * i + a + b.

Or you could simplify the notation by using

Code:

       scanf("%s%d%f",ptr[i].c,&ptr[i].a,&ptr[i].b);
...
       printf("%s\t%d\t%.2f\n",ptr[i].c,ptr[i].a,ptr[i].b);