question about pointers
Author
I have a question on pointers. I have been reading a book on c++,and it was showing me alittle about overloading functions. I understand the overloading part of it, but there's only one thing I don't understand.On one of the overloaded functions it has a ** to one of the parameters.Whats the reason for that? Her is the code.
//code begin
// print_arrs.cpp
#include <iostream>
using namespace std;
void print_arr(int *arr, int n);
void print_arr(double *arr, int n);
void print_arr(char **arr, int n);
int a[] = {1, 1, 2, 3, 5, 8, 13};
double b[] = {1.4142, 3.141592};
char *c [] = {"Inken", "Blinken", "Nod"};
int main()
{
print_arr(a, 7);
print_arr(b,2);
print_arr(c, 3);
return 0;
}
void print_arr(int *arr, int n)
{
for (int i = 0; i < n; i++)
cout << arr << " " ;
cout << endl;
}
void print_arr(double *arr, int n)
{
for (int i = 0; i < n; i++)
cout << arr << " " ;
cout << endl;
}
void print_arr(char **arr, int n)
{
for (int i = 0; i < n; i++)
cout << arr << " " ;
cout << endl;
}
// code end
void print_arr(int *arr, int n);
It's a function that take an int* arr and an int n.
Arguments:
int n: It's easy to understand only an int.
int *arr: It's a pointer to int, you pass the int's address memory
void print_arr(char **arr, int n);
It's a function that take an int *arr and an int n
int n: It's esasy to understand, only an int.
int **arr: It's a pointer to an int pointer, you pass the POINTER TO INT's adress memory.
Example:
int* a = 0;
int j = 6;
int n = 5;
a = j; // A points to j
// First Function
print_arr(a, n); // 1st argument: you pass the int's address memory pointed by a
// j adress memory
// Second Function
print_arr(&a, n); // 1st argument: you pass the POINTER TO INT's adress memory
// a adress memory
One adress is for a int's memory location and the other adress is for pointer to int's memory location
It's a function that take an int* arr and an int n.
Arguments:
int n: It's easy to understand only an int.
int *arr: It's a pointer to int, you pass the int's address memory
void print_arr(char **arr, int n);
It's a function that take an int *arr and an int n
int n: It's esasy to understand, only an int.
int **arr: It's a pointer to an int pointer, you pass the POINTER TO INT's adress memory.
Example:
int* a = 0;
int j = 6;
int n = 5;
a = j; // A points to j
// First Function
print_arr(a, n); // 1st argument: you pass the int's address memory pointed by a
// j adress memory
// Second Function
print_arr(&a, n); // 1st argument: you pass the POINTER TO INT's adress memory
// a adress memory
One adress is for a int's memory location and the other adress is for pointer to int's memory location
'*' means a pointer to, or conventionally, the address of. I single '*' means a pointer to or address of the variable supplied. It could also be the start of an array of such variables.
'**' means the same thing, but once removed. Its a pointer to a pointer, or clearer: a pointer to an address of a variable. It could also be the start of an array of addresses for a given variable type. This means 'char **p' makes p a pointer to a char*, or an array of char* values.
Does this make sense?
One reason you might pass a char** to a function is so you can do something like this:
void print_arr(char **arr, int n);
char bob[12][6];
print_arr(bob, 17);
See? You pass 2D arrays as **, 1D arrays as *.
void print_arr(char **arr, int n);
char bob[12][6];
print_arr(bob, 17);
See? You pass 2D arrays as **, 1D arrays as *.
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