Hai yifangt

allocate memory like this, for your program

Code:

char *text=(char*)malloc(64);
char *a=(char*)malloc(64);

char a[7] = "HELLO";

Here String "Hello" is stored in Character Array 'a', where size is 7. here array 'a' stores characters in contiguous Memory Location. It will take following form after initialization.Each array location will get following values - and last 2 are unused.

Code:

| 0 | 1 | 2 | 3 | 4 |  5 | | |
|'H'|'E'|'L'|'L'|'O'|'\0'| | |

In this method element is accessed Sequentially

char *a = "HELLO";

String "Hello" will be stored at any Anonymous location in the form of array. We even don't know the location where we have stored string, However String will have its starting address.

here Address = [Base Address of Anonymous Array] + [i]

suppose you have to access a[3] then you need to get pointer value, add 3 to the pointer value, and gets the character pointed to by that value.

Code:

#include <stdio.h>

int main()
{
  char (*ptr)[7];
  char arr[7] = {'y','i','f','a','n','g','t'};
  int i;

  ptr = &arr;
  for(i=0; i<7; i++)
  {
    printf("Pointer value: %c Array value: %c\n", (*ptr)[i],arr[i]);
  }
}

Code:

$ ./a.out 
Pointer value: y Array value: y
Pointer value: i Array value: i
Pointer value: f Array value: f
Pointer value: a Array value: a
Pointer value: n Array value: n
Pointer value: g Array value: g
Pointer value: t Array value: t