Pointer arithmetic to change array values

Question

This is a follow on question to an earlier one.

I have learned some errors of my ways, but have extra questions. My objective is to have a local array in one method changed from another method without the use of any global variables.

void methodOne(){

        int myArray[10] = {0};
        int *pMyArray = myArray;
        
        methodTwo(&*pMyArray);
}

This should be declaring an array of null values and passing a reference to the second array as I was shown how to do so correctly here.

void methodTwo(int *passedPointer){
       
        int *localPointer =  passedPointer;

}

Next I'd like to change the values of myArray from methodTwo. So to change the first [0] element I would say:

*localPointer  = 1;

Is this correct?

Then to change the next element would I increment the pointer using:

localPoint++;
*localPointer = 2;

Would this change the second value in myArray? I'm not sure thats the correct way to do it is it?

TIA

What happened when you tried it?

Oliver Charlesworth
– Oliver Charlesworth

2012-04-23 16:22:08 +00:00
Commented Apr 23, 2012 at 16:22 — Oliver Charlesworth
– Oliver Charlesworth, Commented Apr 23, 2012 at 16:22
Just access with localPointer[i].

Daniel Fischer
– Daniel Fischer

2012-04-23 16:22:32 +00:00
Commented Apr 23, 2012 at 16:22 — Daniel Fischer
– Daniel Fischer, Commented Apr 23, 2012 at 16:22

Jerry Coffin · Accepted Answer · 2012-04-23 16:24:16Z

1

Yes, this all looks like it should work. Generally when you're passing an array to a function, you don't assign to a local variable first though -- you just pass the name of the array as the parameter: methoTwo(myArray);. The compiler will automatically convert the name of the array to a pointer to the beginning of the array.

Also note that you can use array-style notation on the receiving end, something like:

localpointer[0] = 1;
localpointer[1] = 2;

...is also reasonable and will accomplish the same as your

*localpointer = 1;
++localpointer;
*localpointer = 2;

For what it's worth, another alternative would be:

*localpointer++ = 1;
*localpointer = 2;

answered Apr 23, 2012 at 16:24

Jerry Coffin

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2 Comments

James MV Over a year ago

I never knew you could use the pointer with array notation. That makes it much easier. Ty.

Jerry Coffin Over a year ago

@EricBanderhide: Yup -- C says that *(pointer+offset) is equivalent to pointer[offset].

Michael Wild · Accepted Answer · 2012-04-23 16:26:08Z

0

First, you pass the array incorrectly. An array is nothing more than a pointer to the first, element, so you don't need to take it's address to pass it to MethodTwo. However, what you should do, is pass in the number of elements in that array to MehtodTwo:

 methodTwo(MyArray, 10);

Since an array is nothing other than a pointer to the first element, you can directly index into the array without requiring the use of pointer arithmetics:

void methodTwo(int passedArray[], size_t n)
// alternative void methodTwo(int *passedArray, size_t n)
{

  passedArray[0] =  passedPointer;

}

answered Apr 23, 2012 at 16:26

Michael Wild

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Comments

Puppy · Accepted Answer · 2012-04-23 16:30:38Z

0

No, this is wrong. Very wrong. Array to pointer decay is really, really bad and you should never, ever, use it. Use a class-based wrapper like std::array, and pass in a pair of iterators to MethodTwo.

answered Apr 23, 2012 at 16:30

Puppy

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