While using a pointer to a 2D array to print any column out of it. How does the pointer point to the-CodePudding

int a[NUM_ROWS][NUM_COLS], (*p)[NUM_COLS], i=5;
//pointer can point to an array of length 'Length Columns'

Let's iterate it

for(p=&a[0]; p<&a[NUM_ROWS]; p  ){

  (*p)[i]=0;

}

My (incomplete) understanding is that p is pointing to the location of array ' a's ' 0th index location, this is stored as a 1-D array of length [NUM_COLS]. How is this making the array column 'i' to reset itself to 0. How is pointer jumping to next column location?

I know that a 'p i' refers to 'address of a 4*i bytes' so how is 'address of a x bytes' happening in column wise iteration using pointer to an array,

CodePudding user response：

p is a pointer to a whole row of the array, so p makes it point to the next row. By setting the [i] entry to 0 in every row, you set the entire i-th column to 0.

I know that a 'p i' refers to 'address of a 4*i bytes' so how is 'address of a x bytes' happening in column wise iteration using pointer to an array,

This is not true.

CodePudding user response：

Instead of this expression in the for loop

p=&a[0];

you could just write

p = a;

Array designators used in expressions are implicitly converted (with rare exceptions) to pointers to their first elements.

So dereferencing the pointer *p you will get the element of the two-dimensional array pointed to by the pointer p. In case of the array a elements of the array has the type int[NUM_COLS].

In turn the array designator *p used in the expression (*p)[i] also is converted to pointer to its first element of the type int *. And the expression (*p)[i] yields the i-th element of the one-dimensional array pointed to by the pointer p.

The for loop can look like

for( p = a; p < a   NUM_ROWS; p   ) 
{
  (*p)[i]=0;
}

That is the i-th element of each "row" of the two dimensional array is set to 0.

To make it more clear consider a one dimensional array.

int a[NUM_ELEMENTS};

then a pointer to the first element of the array looks like

int *p1 = a;

A pointer to the whole array as one object looks like

int ( *p2 )[NUM_ELEMENTS] = &a;

And you may write

p1 = *p2;

because the expression *p2 yields the array a.

As for your statement

I know that a 'p i' refers to 'address of a 4*i bytes' so how is 'address of a x bytes' happening in column wise iteration using pointer to an array,

then actually the expression p i refers to the i-th element of the two dimensional array the offset of the address of which is equal to the value i * sizeof( int[NUM_COLS] ) that is equivalent to i * NUM_COLS * sizeof( int ).