Question

Let’s say we havew a function that returns an array.

int *theFunction()
{
int a[]={1,2,3,4,5};
return a;
}

I want to store the result of the function in a pointer.

int *a=theFunction();

How do I print the array, is it even possible?
The length of the array isn’t known, is there a way to find it?

Answer 1

You cant print an array when its size is unknown

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You can get around this problem by

1>providing a sentinal value i.e the value that would denote the end of the array.
Let us consider -1 to be the sentinal then

int a[]={1,2,3,4,5,-1};
                    ^
                    |->-1 would denote the end of array

But this solution would fail if one of your value seems to be -1

OR

2>Use a struct to denote the size and the array content

struct list
{
int *values;//the array of values
int size;//size of the values array
}

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Also dont return a pointer to value that has local scope..i.e in your case the local value is a..As soon as you get out of the method a is destroyed..further access to this a after the method returns is an error