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.
There are no array operations in C. Arrays implicitly convert to pointers, thus any operation you might attempt upon an array you would actually perform on a pointer.
constant pointer and character pointer
By returning a pointer to the first element of the array.
Mentioning the array name in C or C++ gives the base address in all contexts except one. Syntactically, the compiler treats the array name as a pointer to the first element. You can reference elements using array syntax, a[n], or using pointer syntax, *(a+n), and you can even mix the usages within an expression. When you pass an array name as a function argument, you are passing the "value of the pointer", which means that you are implicitly passing the array by reference, even though all parameters in functions are "call by value". There is, however, one very important distinction. While an array name is referentially the same as a pointer, it is not a pointer in that it does not occupy program referential space in the process. This means that, while you can change the value of a pointer, and thus the address to which it points, you can not change the value of an array name. This distinction is what we call R-Value (array or pointer) as opposed to L-Value (pointer only), i.e. can the object appear on the left sign of an assignment operator.
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.
There are no array operations in C. Arrays implicitly convert to pointers, thus any operation you might attempt upon an array you would actually perform on a pointer.
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 ...}
We use a pointer to reference a string because a string is an array of characters where every element is a char (or a wchar_t if using UNICODE strings). Passing arrays by value would require the entire array to be copied, but passing a pointer variable to an array only copies the pointer, which is effectively the same as passing the array by reference. #include <iostream> int main() { char * psz = "hello"; // pointer to a null-terminated string. std::cout << psz; // pass the pointer (by value) to the insertion operator. return( 0 ); }
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.
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.