Assigning values to an array of struct (or class) behavior

Question

From Java I'm used that I can do this:

public class SomeClass {
    public int field1;
}

SomeClass[] data = new SomeClass[10];
for (int i = 0; i < 10; i++) {
    data[i] = new SomeClass(); 
    SomeClass d = data[i];
    d.field1 = 10;
}

for (int i = 0; i < 10; i++) {
    SomeClass d = data[i];
    System.out.println("Value of data [" + i + "] is [" + d.field1 + "]");
}

And this will work fine, printing:

Value of data [0] is [10]
Value of data [1] is [10]
... etc

So in Java you first create the Array, which has by default all null values, and then in the first loop I create a new SomeClass and assign it to a slot in the array. If you don't do this you get a NullPointerException.

All fine.

The question: I try to do the same in C++. I have one failing and one working example. But I am unsure why it works (or does not work, depending what example you pick). Can anyone elaborate?

First, the failing code:

int main(int argc, char **argv) {
    MyClass data[10];
    for (int y = 0; y < 10; y++) {
        MyClass d = data[y];
        cout << "INITIALLY (garbage): [" << d.field1 << ", " << d.field2 << "] " << endl;
        // assign values
        d.field1 = 2;
        d.field2 = 3;
    }

    // print out data of first 10
    cout << "Printing out data after initialization" << endl;
    for (int y = 0; y < 10; y++) {
        MyClass d = data[y];
        cout << "[" << d.field1 << ", " << d.field2 << "] " << endl;
    }
}

So If i understand correctly, according to this StackOverflow question I can create an array of a Struct like I do in the above code.

What I noticed is that if I do not use:

MyClass d = data[y];
d.field1 = 2;
d.field2 = 3;

But instead I do:

data[y].field1 = 2;
data[y].field2 = 3;

It does work.

However, if I insist on using a separate value, I can still make it work by doing this:

MyClass * d = &data[y];
d->field1 = 2;
d->field2 = 3;

I do not change anything in printing the output. And the above works.

So, something clearly is different when using a pointer to data[y]. I cannot found a clear answer on this though. Anyone able to explain why?

If this question is a duplicate, sorry for that, I could not find a real answer on the "why" part. Code snippets are not always enough for me ;)

PPS: I am aware that I am not allocating this array in the heap. Bonus points for touching that subject to compare to :)

ecatmur · Accepted Answer · 2013-02-26 12:07:22Z

1

In C++, an object is a region of storage - that is, an area of memory. Variables denote objects directly, unless they are explicitly qualified as pointers or references to existing objects.

When you write:

MyClass d = data[y];

you are declaring a variable of type MyClass, defining it as denoting an object on the stack, and initialising it from the object data[y]; d and data[y] are separate objects so modifying d will not affect data[y].

You want to write:

MyClass &d = data[y];
        ^--- here

The ampersand & denotes d as a reference to the object data[y], so as long as an object continues to exist within the region of storage that data[y] originally denoted (i.e. at that address), you can use d to modify that object.

Using a pointer is similar, but because pointers can be reseated to point to a different object you must use indirection syntax (* or ->) to denote the object they point to.

answered Feb 26, 2013 at 12:07

ecatmur

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1 Comment

Stefan Hendriks Over a year ago

Wow thanks a bunch! Really clear answer. Do you have any further reference for me to get a better understanding of these behaviours? For example, this is a one dimensional array. What If I'd like to have a two dimensional array? In Java things are just much easier to grasp. But I insist on doing this in C++. Again, thanks a lot!

John Zwinck · Accepted Answer · 2013-02-26 12:01:59Z

1

This:

    MyClass d = data[y];
    d.field1 = 2;
    d.field2 = 3;

Has no effect because d is now a copy of data[y]. If you said MyClass& d instead, you'd have a (non-const) reference, which you could modify and then observe those changes in data.

answered Feb 26, 2013 at 12:01

John Zwinck

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Comments

juanchopanza · Accepted Answer · 2013-02-26 12:06:49Z

1

You are making a copy here:

MyClass d = data[y];

Modifications to d have no impact on data[y].

Take a reference instead:

MyClass& d = data[y];

Now d refers to the element data[y]. Note that in C++ a reference behaves like an alias. You cannot re-assign it. For example

d = someMyClassInstance;

will have the effect of assigning the value of someMyClassInstance to data[y].

edited Feb 26, 2013 at 12:06

answered Feb 26, 2013 at 12:01

juanchopanza

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1 Comment

Stefan Hendriks Over a year ago

Thanks a lot! I did not realize I was creating a copy. It makes sense now.

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