CODEWITHSUNDEEP
c
redsoxlost
redsoxlost

Reputation: 1235

Dynamic Memory Allocation - When and Why

I am a novice in C and trying to understand basic concepts in C, Like when and Why I have to allocate Memory for a string pointer. here I have a sample program. I have commented in appropriate locations. Please help me understand. 

/ Please help me understand why do I have allocate Memory in Case 2 while I don't have to in Case 1.

#include<stdio.h>
void xcopy(char *t,const char *s);
int  main(int argc, char const *argv[])
{
  char name1[]="Asfakul";
  char *name;
  char *target;
  name=name1; // Here I dont have to allocate Memory (Case 1)
  puts(name);

  target=(char*)calloc(10,sizeof(char));  // Here I have to allocate Memory (Case 2)
  xcopy(target,name);
  return 0;
}


void xcopy(char *t,const char *s)
{
  while( *s !='\0')
  {
    *t=*s;
    t++;
    s++;

  }
  puts(t);

}

Upvotes: 0

Views: 61

Answers (2)

dbush
dbush

Reputation: 223699

In the first case, you start with name1, which is an array of char. Then you take name, which is a char *, and assign name1 to it. Since name1 is being evaluated in pointer context, it refers to a pointer to the first element of the array. So now name points to the first element of name1.

In the second case, target is assigned a memory location returned by a call to calloc, which in this case is a block of 10 bytes. The bytes now pointed to by target can now be used.

As with any pointer, you need to assign it a value before you can dereference it. That value can be either the address of some other variable, or a block of memory returned by the malloc family of functions.

Upvotes: 1

ameyCU
ameyCU

Reputation: 16607

name=name1; // Here I dont have to allocate Memory (Case 1)

In this case , you don't allocate memory to name , you just make it point to array name1 . In short , name now has address of first element of array name1. 

target=(char*)calloc(10,sizeof(char));  // Here I have to allocate Memory (Case 2)
xcopy(target,name);

And in case 2 you need to allocate memory as you copy the contents of name1 at the memory block to which target points to .

Here , it is needed as if you don't allocate memory then target point to anything (maybe garbage) and writing to that location will cause undefined behaviour.

Note - You don't need to free any memory in case 1 as pointer points to a array on stack . But you need to do free(target); in case 2 as you allocate memory on heap.

Upvotes: 1

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