Thread: Array question

To pass an array to a function you could do this:

Code:

#include<iostream>
using namespace std;

void someFunc(const int* array, int n)
{
for (int i=0;i<n;i++)
  cout<<array[i]<<endl;
}

int main()
{
int array[10];
for (int i=0;i<10;i++)
  array[i]=i;
someFunc(array,10);
}

In that example you could not modify the contents of the array and n would be the size of the array. Does that help?

Whenever you pass an array, you pass a pointer. Regarding functionality, the function declarations below are the same:

Code:

void f( int* array_in, int size_in ); // pass a pointer - actually it should point to the first element of the array 

void g( int array[], int size_in ); // pass an array

Both function have a second parameter, as you should also provide the array's size, in order to prevent step-by errors - meaning you try to read or write data pass the end of the array. Passing the array's size allows necessary safety checks which are recommended when working with arrays.

When calling the function, you can do following:

Code:

int anArray[5] = {0, 1, 2, 3, 4}; // declaration and initilization
f( &anArray[0], 5 ); // passing the address of the first element

f( anArray, 5 ); // simplified syntax

Using the second, simplified form is shorter and safer, as you surely pass the starting address of the given array (the compiler ensures that).
As arrays are passed by pointers, not by value, their contents can be changed in the called function. In order to disallow this, you should pass a pointer to constant, as specified by JaWib.

Speaking of classical arrays, their size does not change at runtime. Of course, you can write your own Array class, which does change dinamically, but you should also check STL vector class for example. However, this is a little bit advanced issue.