Relationship between C ++ arrays and pointer types

1.

The following array definitions are known.
Int Ia [] = {0, 1, 1, 2, 3, 5, 8, 13, 21 };
Therefore, it is easy to write.
IA;
What does it mean?
The array identifier represents the address of the first element in the array. Its type is the pointer of the array element type. In Ia, its type is int *. Therefore, the following two forms are equivalent. returns the address of the first element of the array.
IA;
& Ia [0];

Similarly, to access the corresponding value, we can take one of the following two methods:
// Both get the value of the first element
* IA;
IA [0];
We know how to use the subscript operator to access the address of the second element.
& Ia [1];

Similarly, the following expression
IA + 1;

The address of the second element can also be obtained. Similarly, the following two expressions can access the value of the second element.
* (IA + 1 );
IA [1];

But the following expression

* Ia + 1; // The unreferenced operator has a higher priority than the addition operator. The result is to add 1 to the first element.
It is completely different from the following expression.
* (IA + 1); // The bracket operator has a higher priority than the unreferenced operator. The result is the value of the second element.

 

2.

Traverse arrays through pointers

# Include <iostream>
 
Template <class elemtype> // Template Function
Void print (elemtype * pbegin, elemtype * pend)
{

// The pend pointer executes the next bit of the last element
While (pbegin! = Pend ){
Cout <* pbegin <'';
++ Pbegin;
}
}

// Main Function

Int main ()
{
Int Ia [9] = {0, 1, 1, 2, 3, 5, 8, 13, 21 };
Double da [4] = {3.14, 6.28, 12.56, 25.12 };
String SA [3] = {"Piglet", "Eeyore", "Pooh "};
Print (IA, Ia + 9); // The second parameter Pointer Points to the next position of the last element of the array
Print (Da, da + 4); // da + 3 + 1
Print (SA, Sa + 3); // SA + 2 + 1
}