CODEWITHSUNDEEP
c++
James MV
James MV

Reputation: 8717

Pointer arithmetic to change array values

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

Upvotes: 0

Views: 10111

Answers (3)

Puppy
Puppy

Reputation: 146930

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.

Upvotes: 0

Michael Wild
Michael Wild

Reputation: 26341

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;

}

Upvotes: 0

Jerry Coffin
Jerry Coffin

Reputation: 490148

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;

Upvotes: 1

Related Questions