Void reinterpret cast question

[jfsd]

Hello, I have got a question, in the following code:

void func(void* v, int n, int val) {
    int* xp = reinterpret_cast<int*>(&v);
    for(int* i=xp; i<xp+n; i++)
       *i = 5;
}
int main() {
    int array[10];
    func(array, 10, 5);

    for (int j=0; j<10; j++)
        cout << array[j] << endl;
}

When passing the array into the function the array values don't change. However, if I just put the function contents directly in main(), I manage to change the values of the array.
Am I doing something wrong?

[newb16]

because this is not original pointer to the beginning of array, but pointer to
v pointer itself ( &v ).
int* xp = reinterpret_cas t<int*>(&v);

[jfsd]

Thanks for the hint. Changing the line to this makes it work:

int* xp = reinterpret_cast<int*>(v);

[weaknessforcats]

reinterpret_cas t is not required.

What is required is understanding how arrays work.

Read this: http://bytes.com/forum/thread772412.html.

I have never seen a case in C++ where you need write a new function with a void* argument.

[Billiska]

First, I agree with the above reply that we rarely need to use void pointer as an argument.
However, I still think there are some circumstances that we need it.

Anyway, for this case, why don't you change it to

void func(int* v, int n, int val) {

or

void func(int v[], int n, int val) {

instead, so that you don't have to worry about changing the type.

Another alternative way would be to use "memset" function.
You can find a documentation here

And if you want to use this function for many types of data, I suggest you define it using "template " like this:

template <class T>
void func(T* v, T n, T val) {
    T* xp = v;
    for(T* i=xp; i<xp+n; i++)
       *i = val;
}

*I think you want to put "val" there instead of "5", so I changed it.