Individual elements of an array as a function parameters in C?-CodePudding

Suppose I have a function foo:

int foo(int a, int b)
{
    return a   b;
}

Can I in any way call foo with an array and have each elements of the array act as one parameter?

e.g:

int arr[2] = {1, 2};
foo(arr); // Should return 3

In JavaScript I can do:

let arr = [1, 2];
foo(...arr); // Returns 3

Is there anything similar in C?

CodePudding user response：

No, this is not possible. You will have to call the function foo like this:

int arr[2] = { 1, 2 };
foo( arr[0], arr[1] );

However, it is possible to redefine the function foo like this:

int foo( int arr[2] )
{
    return arr[0]   arr[1];
}

Now, you can call the function foo like this:

int arr[2] = { 1, 2 };
foo( arr );

CodePudding user response：

You can pass an array to a function, but in doing so it decays to a pointer so you don't know what the size is. You would need to pass the size as a separate parameter, then your function can loop through the elements.

int foo(int *arr, int len)
{
    int i, sum;
    for(i=0, sum=0; i<len; i  ) {
        sum  = arr[i];
    }
    return sum;
}

Then you can call it like this:

int arr[2] = {1, 2};
foo(arr, 2);

Or, assuming the array was declared locally:

foo(arr, sizeof(arr)/sizeof(*arr));

CodePudding user response：

Only by writing out the references longhand:

foo(arr[0], arr[1]);