copy array in c++

Question

Hi, I've got this function:

void vector(int imputC[]) { 
    //return v*   
    //imputC[].lenght
    int a =sizeof(imputC) / sizeof(int);
    // float v[2 * a + 2];
    // v[0]=0;
    for (int i = 0; i < a ; i++)
        v[i] =(float)imputC[i];

    for(int j = 0; j < 10; j++)
        cout << a << endl;
}

And main:

int main() {
    int array[] = { 23, 5, -10, 0, 0, 321, 1, 2, 99, 30 };
    // float *v[22];
    vector(array);
}

And size of a is 1. Why is that? I don't understand, I think that's is the only object.

[sizeof()](http://en.wikipedia.org/wiki/Sizeof) is not for that. *"sizeof is used to calculate the size of any datatype, measured in the number of **bytes required to represent the type**"*. Read the documentation before using a function. — m0skit0, Jan 10 '13 at 17:23
@MarcB - just to really really make sure that a is 1. That, or to make the array have 1111111111 elements. — slugonamission, Jan 10 '13 at 17:24
Assuming you're actually using C++, just use a vector instead of an array, and copying becomes trivial (just use an assignment like `auto b = a;`). — Jerry Coffin, Jan 10 '13 at 17:24

score 3 · Answer 1 · answered Jan 10 '13 at 17:22

The expression sizeof(imputc) doesn't work in your function, as the array is passed as a pointer and doesn't retain the size of the original array. So that expression returns only the size of the pointer not the array it points to.

If you must pass an array to a function that needs to know the number of items in the array, you must always pass the array size as an argument to that function.

score 1 · Answer 2 · answered Jan 10 '13 at 17:23

In a declaration of a function that takes an array type as an argument the array type decays into a pointer to the first element. So

void vector(int imput[])

is the same as

void vector(int *imput)

A function that takes an array argument needs to have some other way of getting the number of elements in the array.

copy array in c++

2 Answers2