C++ Passing array into function

Question

I'm relatively new to C++ and am having a tough time passing my array into a separate function. Apologies for re-asking a question that has no doubt been answered a dozen times before, but I couldn't find any questions similar to the problem I have with my code.

int main()
{
    Array<int> intarray(10);
    int grow_size = 0;

    intarray[0] = 42;
    intarray[1] = 12;
    intarray[9] = 88;

    intarray.Resize(intarray.Size()+2);
    intarray.Insert(10, 6);

    addToArray(intarray);

    int i = intarray[0];

    for (i=0;i<intarray.Size();i++) 
    cout<<i<<'\t'<<intarray[i]<<endl;

    Sleep(5000);
}

void addToArray(Array<int> intarray)
{
    int newValue;
    int newIndex;

    cout<<"What do you want to add to the array?"<<endl;
    cin >> newValue;
    cout<<"At what point should this value be added?"<<endl;
    cin >> newIndex;

    intarray.Insert(newValue, newIndex);
}

You need to provide the Array::Insert implementation if you want someone to help you. — Kevin
– Kevin, Commented Jun 28, 2013 at 15:15
@JamesMcLaughlin It is new in C++11 cplusplus.com/reference/array/array — stonemetal
– stonemetal, Commented Jun 28, 2013 at 15:16
@stonemetal That's std::array, not Array as used in this program. — bames53
– bames53, Commented Jun 28, 2013 at 15:17
Ahh, I shouldn't have deleted my comment, then. So what is this mysterious Array? — James M
– James M, Commented Jun 28, 2013 at 15:17

trojanfoe · Accepted Answer · 2013-06-28 15:13:56Z

5

You are passing a copy of the array, so any changes will not affect the original. Pass by reference instead:

void addToArray(Array<int> &intarray)
//                         ^

answered Jun 28, 2013 at 15:13

trojanfoe

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5 Comments

David G Over a year ago

That's not his only problem though.

trojanfoe Over a year ago

@0x499602D2 Care to elaborate?

David G Over a year ago

He declared an object but he's using it as an array: Array<int> intarray(10). That should be Array<int> intarray[10]

trojanfoe Over a year ago

@0x499602D2 I'm not sure that's true; the constructor could be Array::Array(size_t size), which seems reasonable. Without seeing Array.h though, I couldn't say for sure.

Liam Over a year ago

Thanks, that solved the problem I was having. I also realized that I had forgotten to declare the function. It seems to be working fine now.

Mr.C64 · Accepted Answer · 2013-06-28 16:59:56Z

This is a special case of a more general question on parameters passing.

You may want to consider the following guidelines:

If you want to pass something to a function to modify it inside the function (and make the changes visible to the caller), pass by reference (&).

e.g.

// 'a' and 'b' are modified inside function's body,
// and the modifications should be visible to the caller.
//
//     ---> Pass 'a' and 'b' by reference (&) 
//
void Swap(int& a, int& b)
{
    int temp = a;
    a = b;
    b = temp;
}

If you want to pass something that is cheap to copy (e.g. an int, a double, etc.) to a function to observe it inside the function, you can simply pass by value.

e.g.

// 'side' is an input parameter, "observed" by the function.
// Moreover, it's cheap to copy, so pass by value. 
//
inline double AreaOfSquare(double side)
{
    return side*side;
}

If you want to pass something that is not cheap to copy (e.g. a std::string, std::vector, etc.) to a function to observe it inside the function (without modifying it), you can pass by const reference (const &).

e.g.

// 'data' is an input parameter, "observed" by the function.
// It is in general not cheap to copy (the vector can store
// hundreds or thousands of values), so pass by const reference.
//
double AverageOfValues(const std::vector<double> & data)
{
    if (data.empty())
        throw std::invalid_argument("Data vector is empty.");

    double sum = data[0];
    for (size_t i = 1; i < data.size(); ++i)
        sum += data[i];

    return sum / data.size();
}

In modern C++11/14 there is also an additional rule (related to move semantics): if you want to pass something that is cheap to move and make a local copy of it, then pass by value and std::move from the value.

e.g.

// 'std::vector' is cheap to move, and the function needs a local copy of it.
// So: pass by value, and std::move from the value.
//
std::vector<double> Negate(std::vector<double> v)
{
    std::vector<double> result( std::move(v) );
    for (auto & x : result)
        x *= -1;
    return result;
}

Since in your addToArray() function you modify the Array<int> argument, and you want modifications visible to the caller, you can apply rule #1, and pass by reference (&):

void addToArray(Array<int> & intarray)

Adib Bazli · Accepted Answer · 2017-06-07 04:05:50Z

-1

My suggestion is use pointer, as usual declare the function

void yourFunction(int * arrayPointer);

In main() function you should input

int yourArray[8] = {11, -2, 45, 37, 18, 35};
int * arrayPointer = &yourFunction[0];
yourFunction(yourArray); // call your function

Finally in yourFunction()

void yourFunction(int * arrayPointer)
{
    // print all number in the array
    for(int x = 0; x < 6; x++)
    {
        cout << *arrayPointer << " ";
        arrayPointer++;
    }
}

This at least work for me, hope this will help

answered Jun 7, 2017 at 4:05

Adib Bazli

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