I have an array int arr[5]
that is passed to a function fillarr(int arr[])
:
int fillarr(int arr[])
{
for(...);
return arr;
This:
int fillarr(int arr[])
is actually treated the same as:
int fillarr(int *arr)
Now if you really want to return an array you can change that line to
int * fillarr(int arr[]){
// do something to arr
return arr;
}
It's not really returning an array. you're returning a pointer to the start of the array address.
But remember when you pass in the array, you're only passing in a pointer. So when you modify the array data, you're actually modifying the data that the pointer is pointing at. Therefore before you passed in the array, you must realise that you already have on the outside the modified result.
e.g.
int fillarr(int arr[]){
array[0] = 10;
array[1] = 5;
}
int main(int argc, char* argv[]){
int arr[] = { 1,2,3,4,5 };
// arr[0] == 1
// arr[1] == 2 etc
int result = fillarr(arr);
// arr[0] == 10
// arr[1] == 5
return 0;
}
I suggest you might want to consider putting a length into your fillarr function like this.
int * fillarr(int arr[], int length)
That way you can use length to fill the array to it's length no matter what it is.
To actually use it properly. Do something like this:
int * fillarr(int arr[], int length){
for (int i = 0; i < length; ++i){
// arr[i] = ? // do what you want to do here
}
return arr;
}
// then where you want to use it.
int arr[5];
int *arr2;
arr2 = fillarr(arr, 5);
// at this point, arr & arr2 are basically the same, just slightly
// different types. You can cast arr to a (char*) and it'll be the same.
If all you're wanting to do is set the array to some default values, consider using the built in memset function.
something like: memset((int*)&arr, 5, sizeof(int));
While I'm on the topic though. You say you're using C++. Have a look at using stl vectors. Your code is likely to be more robust.
There are lots of tutorials. Here is one that gives you an idea of how to use them. http://www.yolinux.com/TUTORIALS/LinuxTutorialC++STL.html