CODEWITHSUNDEEP
c++arrayspointersreference
Pipsydoodles
Pipsydoodles

Reputation: 37

Having difficulty working with pointers

I having some issue when it comes to initializing pointers.

void findMM (int *PMM, int *theG)
{
        // code  I haven't written yet. It will essentially take two variables from   //theG and store it in MM  
}

int main() 
{ 
    int size;
    int MM [2] = {1000, 0}; 
    int *theG = NULL; 
    cout << "\nPlease insert size of array:" << endl; 
    cin >> size; 
    theG = new int [size];  
    findMM(&MM, &theG); //Get error with &MM  
    delete [] theG;     
    return 0; 
}

The complier says that argument of type int (*)[2] is incompatible with parameter of type int ** So obviously that I have issue with the code in particular my (reference?) of array MM. Or perhaps there is other obvious faults that I am missing?

Edit attempt 2 

void findMM (int *PMM, int *theG)
{
        PMM [1] = 5; 
        theG [0] = 7; 
}

int main() 
{ 
    int size;
    int MM [2] = {1000, 0}; 
    int *theG = NULL; 
    cout << "\nPlease insert size of array:" << endl; 
    cin >> size; 
    theG = new int [size];  
    findMM(MM, theG);
    cout << MM [1] << endl << theG[0];    
    delete [] theG;     
    return 0; 
}

The output would be 5 and 7 correct?

Upvotes: 0

Views: 143

Answers (2)

danieltm64
danieltm64

Reputation: 481

I know this question has already been answered but I believe I can contribute to the asker's understanding.

Let's start with the basics:

void main()
{
    int a = 2; // a is an int
    cout << a << endl; // print 2

    int *b; // b is a pointer-to-int
    b = &a; // store the address of a in b
    cout << *b << endl;// print the value that b points to, which is 2

    int my_array = new int[3]; // allocate an array with 3 integers
    my_array[0] = 50; // store 50 in the first element of the array
    my_array[1] = 51; // store 51 in the second element of the array
    my_array[2] = 52; // store 52 in the third element of the array
    cout << c[0] << endl; // print 50

    some_function(my_array, 3); // explained below
}

Now let's see how to pass arrays into functions. Assume we want to have a function called some_function that receives an array.

void some_function(int *some_array, int size_of_the_array)
{
    // use the array however you like here
}

The function some_function receives a pointer to an int (also known as "pointer-to-int"). The name of an array is always the address of its first element, so if a function expects a pointer to an int and you give it the name of an array of ints, you are actually giving it the address of the first element in the array (this is just C++ syntax rules). So the function now has the address of the first element in the array, it can do stuff like *some_array to access the first element in the array, but what if it wants to access the other elements? It adds 1 to the pointer it already has and then applies the * operator to it: *(some_array + 1). Let's say an int is 4 bytes, if you add 1 to a pointer-to-int, the result of this addition is a new pointer that points to a location in memory 4 bytes ahead, so *(some_array + 93) is the value in the 94th element of the array some_array (array elements are stored sequentially in memory). A shorthand notation for this is some_array[93]. So if you have int *some_array = new int[100];, then some_array is a pointer and some_array[93] is the same as *(some_array + 93), which is the 94th element in the array.

The address itself though is not enough, you also need to know the number of entries in the array so that you don't try to access an element past the end of the array. In this example, assume that some_function simply prints the contents of the array, so if you don't provide 3 as the second argument to the function then it will have no way of knowing when to stop adding 1 to the pointer it received in the first argument. Beware, however, that by passing an array to a function this way, you are not passing the function a copy of the array, you are simply telling it where to find its contents in memory.

Upvotes: 1

Joseph Mansfield
Joseph Mansfield

Reputation: 110658

Since MM is an array, &MM is a pointer to an array (that's the type int (*)[2] that you see in the error). Instead, you seem to want to pass a pointer to the first element of the array. There are two ways to do that. Firstly, you can explicitly get the first element and then take the address of it: &MM[0]. Secondly, you can rely on array-to-pointer conversion to do it for you and just pass MM. Array-to-pointer conversion converts an array to a pointer to its first element.

Upvotes: 3

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