Using malloc in C. Please help me.

**newcprogram** · 06-25-2009

I have a code and i must complete it, make b[i]==a[i].

Code:

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

#define N 5

int a[N]={80,75,50,90,45};

int main(int argc, char **argv) {

int i;

int *b;

b=(int *)malloc(sizeof(int)*N);

if(b==NULL) {
fprintf(stderr,"malloc error\n");
exit(1);
}

for(i=0;i<N;i++) {
 _________________					
     }
 
 for(i=0;i<N;i++) {
     printf("%d\n",b[i]);
 }
 
 free(b);
 }

I must fill in the underline. And i think i must only write:

Code:

for(i=0;i<N;i++) {
      b[i]=a[i];
}

Is that correct?
And when i try to find some example about malloc from internet, i have this code:

Code:

typedef struct data_type {
  int age;
  char name[20];
} data;

data *bob;
bob = (data*) malloc( sizeof(data) );
if( bob != NULL ) {
  bob->age = 22;
  strcpy( bob->name, "Robert" );
  printf( "%s is %d years old\n", bob->name, bob->age );
}
free( bob );

I don't understand why they must write this sign "->"

Code:

bob->age = 22;
  strcpy( bob->name, "Robert" );

thanks for your help.

**Sebastiani** · 06-25-2009

>> And i think i must only write:

That is correct.

>> I don't understand why they must write this sign "->"

Because you're working with a pointer. That's just how it's done. You could of course dereference the pointer, in order to be able to access the members as you would a non-pointer, eg:

Code:

(*bob).age = 22;

>> (int *)malloc(sizeof(int)*N);

You don't need to cast the result. In C, you can assign any kind of pointer directly to a void*.

**sugumar.tr** · 07-25-2009

here I gave one example for this .
:insert

Code:

struct student {
    int age;
    char name[25];
      };

struct student ss;

now the structure variable is "ss" it is not a pointer. I can access the structure member by the following way.
:insert

Code:

ss.age=22;
strcpy(ss.name,"sugumar");
struct student ........;

now the structure variable is pointer variable. I can't access the structure member by the following way. Ss.age=22;

if I want to access the structure member I need to use the "->" symbol.
:insert

Code:

Ss->age=22;
strcpy(Ss->name,"dhivya Lakshmi");

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