doubt in address of array? difference between usage of arr and &arr

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C++

void main()
{
 int arr[]={1,2,3}; 
printf("%u %u",arr,&arr); 
printf("%u %u",arr+1,&arr+1);
}

first statement i am getting same address for arr and &arr.
2nd it gives different y?
wats the difference in using arr and &arr?
can any one explain me?

Posted 29-Jun-15 22:37pm

moyna coder

Updated 29-Jun-15 22:54pm

CPallini

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CPallini · Accepted Answer · 2015-06-29T22:54:00

arr and &arr always hold the same address (the name of an array is a pointer to array first item, also &arr[0] gives the same address).
As for pointer math, the nature (the type) of arr and &arr is different: (arr+1) jumps to the next (the 2^nd) item of the array, while (&arr+1) jumps to the end of the array (the compiler takes into account the array size).

Please note your code should be:

C

#include <stdio.h>

int main()
{
  int arr[]={1,2,3};
  printf("%p %p\n",arr,&arr);
  printf("%p %p\n",arr+1,&arr+1);
  return 0;
}

Quote:
can you say me the use of **p? . its confusing. thankz in advance.

A double pointer could be used, for instance, in jagged array implementation:

C

 #include <stdio.h>
 #include <stdlib.h>
 #define N 3
int main()
{
  int ** pp;
  int size[N] = { 2,3,1 } ;
  pp = (int**) malloc( N * sizeof(int *));

  int n,k, count=0;

  for (n=0; n<N; ++n)
  {
    pp[n] =(int *)  malloc( size[n] * sizeof(int));
    for (k=0; k<size[n]; ++k, ++count)
      pp[n][k] = count;
  }

  for (n=0; n<N; ++n)
  {
    for (k=0; k<size[n]; ++k)
      printf("pp[%d][%d] = %d\n", n, k, pp[n][k]);
    printf("-------------\n");
  }


  for (n=0; n<N; ++n)
  {
    free(pp[n]);
  }
  free (pp);

  return 0;
}