What is the difference between:
//Example of "Complicated Array Declarations" from C++ Primer
int arr[10] = {1,2,3,4,5,6,7,8,9,10};
int (*Parr)[10] = &arr;
And:
int arr[10] = {1,2,3,4,5,6,7,8,9,10};
int *Parr = arr;
Both are pointers to an array of integers. But in order to access the first element of arr in the first snippet, I have to do **Parr whereas in the second, I only have to dereference once *Parr
In the first code sample,
int (*Parr)[10] = &arr;
Parr is a pointer to an array of 10 ints. It can only point to such an object. For example,
int (*Parr)[10];
int a[10];
Parr = &a; // OK
int b[42];
Parr = &b; // ERROR, b is of the wrong type
In the second code sample,
int *Parr = arr;
Parr is a pointer to int, initialized to point to the first element of arr. But it can point to any int.
int* Parr;
int a[10];
Parr = &a; // OK, a decays to int*
int b[42];
Parr = &b; // OK, b decays to int*
int c = 42;
Parr = &c; // OK, c is an int*
(*Parr)[3].
First one is a 'pointer to int[10]'.
int (*Parr)[10] = &arr; // point to int array
*Parr; // *Parr == arr
**Parr; // **Parr == *(arr) == arr[0]
Second one is a 'pointer to int'
int *Parr = arr; // point to the start of array
*Parr; // *Parr == *(arr) == arr[0]
The way we can have a pointer to an integer,or a pointer to a foat ,can we also have a pointer to array? The answer is "YES".Declaration of a pointer to an array,however is a little clumsy.
For Example,The declaration int(*q)[4] means that q is a pointer to an array of 4 integer. in your Code
int(*parr)[10] means parr is a pointer to an array of 10 integer.
but int *parr=arr is only a pointer to the oth element of arr[10].
So suppose you assgin any pointer to arr[10]. in the second case doing parr++.the parr will move to location arr[10] form arr[0]. but in the second case doing parr++.the parr will move to location arr[1] form arr[0].
So i hope u got the answer.
