# REGARDING POINTERS AND MATH ----- ( TUTORIAL 41 )

 saivivek reddy · May 24, 2015 Instead of giving 5 values in the array if some other number of values are given then the addresses are not shown as it was in the video . some one help me with this issue .

## Replies

 Laura Lee · May 24, 2015 I have not seen the video and your question is ambiguous. I assume you're referring to pointer arithmetic and accessing values from an array. `` int arr[10] = {0}; //An array is a static pointer who's position in memory// should not be changed. int* parr = arr; //parr points to the address arr points to arr[0] = 2; // Arbitrary value chosen for the first element of arr printf("%i\n",*parr);//Dereferencing parr will give the value at arr[0]:2 ++parr;//We've increased the position in memory of parr ////because parr is an int array, the position increased four bytes arr[1] = 4;//Arbitrary value printf("%i\n",*parr);//Dereferencing parr will give the value at arr[1] :4 parr+=1;//We can also increase our pointer like this. //Since parr is an int pointer, increasing by n would be n*sizeof(int); arr[2] = 8; printf("%i\n",*parr); //will print 8 parr+=2; //2*sizeof(int) == 8 == two int elements *parr = 32;//Set arr[4] to 32 printf("%i\n",arr[4]);//We skipped past arr[3] ; will print 32``
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## C++

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