A pointer into an array of elements of type E is a pointer to a single element of type E:
typedef ..... E;
E array[123];
E* const pointer = &array[18]; // points to the 19th element inside 'array'
An array of pointers is an array whose elements are pointers:
typedef .... E;
E* array[123];
E** const pointer = &array[18]; // points to the 19th pointer within 'array'
Referencing the name of the array variable without use of the index operator itself is a constant pointer to its first element. Therefore, the following if-clause is always true:
typedef .... E;
E array[123];
if (array &array[N]) { // ALWAYS true ...}
Yes, passing an array name to a pointer assigns the first memory location of the array to the pointer variable. An array name is the same as a pointer to the first location of the array, with the exception that an array name is a r-value, while a pointer is an l-value.
because u freakin can
constant pointer and character pointer
once we initialize the array variable, the pointer points base address only & it's fixed and constant pointer
By returning a pointer to the first element of the array.
In the C and C++ languages the array notation arr[i] is completely equivalent to the pointer notation *(arr + i).
A pointer is a variable that holds address information. For example, in C++, say you have a Car class and another class that can access Car. Then, declaring Car *car1 =new Car() creates a pointer to a Car object.. The variable "car1" holds an address location.
the simple and efficient way to pass an array is pointer to an array like that int (*p)[30] ; // pointer to an array of integer having 30 element
You cannot add elements to a fixed array in C or C++. If, however, the array is declared as a pointer to an array, you can add elements by allocating a new array, copying/adding elements as needed, reassigning the new array to the pointer, and deallocating the original array.
An array behaves like a pointer when you use its name in an expression without the brackets.int a[10]; /* a array of 10 ints */int *b = a; /* a reference to a as a pointer, making b like a */int c = *(a+3); /* a reference to a[3] using pointer semantics */myfunc(a); /* pass a's address, a pointer to myfunc */Note very carefully that, while an array name and a pointer can almost always be interchanged in context, the are not the same, in that a pointer is an l-value, such as b, above, and can be assigned, whereas a is an r-value and can only be referenced, such as in the same statement, the second statement. Also, an array name does not take up memory, while a pointer does.
All variable names are an alias for the value stored at the memory address allocated to them. To get the memory address itself, you must use the address of operator (&). The value returned from this can then be stored in a pointer variable.Arrays are different. The array name is an alias for the start address of the array, thus you do not need the address ofoperator to obtain the memory address (although you can if you want to). This means that when you pass an array name to a function, you pass the memory address of the array rather than passing the array itself (which would require the entire array to be copied, which is a highly inefficient way to pass an array). In essence, the array is passed by reference rather than by value.Consider the following code. This shows how a primitive variable name differs from the name of an array of primitive variables. The final portion shows how a pointer can be used to achieve the same results you got by accessing the array elements directly from the array name itself. This is in fact how the compiler implements arrays, using pointers, but there's no need to do this in your code. Accessing array elements directly by their index is a programming convenience.#include using namespace std;int main(){int i = 10;cout
it tell that the line goes on forever.