'Cannot convert 'float' to 'float*' for argument '1'

Question

I'm an ICT student, studying C++, and I find it very interesting. However, while experimenting, I came up to something I can't fix myself nor find on the net. This is the code:

    #include <iostream>

using namespace std;


float average(float array[10], float &average){
      int i = 0;

      for( i = 0; i != 10; i++ ){
                cout << "Vett ("<<i<<") = ";
                cin >> array[i];
                while(cin.fail())
                {
                              cin.clear();
                              cin.ignore();
                              system("Color 5C");
                              cout << "\nPlease insert a number: ";
                              cin >> array[i];
                }
                average = average + array[i];
                }
                average = average / 10;
                return array[10];
                return average;
                }

main(void){
           float vett[10], media;
           int i;
           char loop;

           vett[10] = 0;
           media = 0;


           do{
           system("cls");
           system("Color 00");
           cout<<"****************************************"<<endl;
           cout<<"***     INSER THE DATA TO COMPUTE    ***"<<endl;
           cout<<"****************************************\n"<<endl;

         /*  for( i = 0; i != 10; i++ ){
                cout << "Vett ("<<i<<") = ";
                cin >> vett[i];
                while(cin.fail())
                {
                              cin.clear();
                              cin.ignore();
                              system("Color 5C");
                              cout << "\nPlease insert a number: ";
                              cin >> vett[i];
                }
                media = media + vett[i];
                }

                media = media / 10;
                */
                average(vett[10],media);
                for( i = 0; i != 10; i++ ){
                     cout << vett[i]<<"  ";
                     }

                if(media == 0){
                         cout << "\nATTENTION the average equals to = "<<media<<endl;
                         }
                else{
                     cout << "\nThe average is"<<media<<endl;
                     }
                     printf("\n");
                     cout << "\nDo You want to continue? Y/N";
                     cin >> loop;
                     }
                     while(loop == 'Y' || loop == 'y');
                     system("pause");
                     }

For some reason I couldn't set in the 'average' function the array as a pointer (&array), perhaps because the array is already a pointer. Nonetheless, removing it gives me the following error:

"Cannot convert 'float' to 'float*' for argument '1' to 'float average(float*,float&)'

If I call the function this way

average(&vett[10],media);

it works, but returns weird values in the array. As you can see, I commented the same thing I put in the function, which works perfectly, unless..I put it in a function. I assume I've done something wrong with the function call, can anybody help me understand?

Solution 1

First of all, note that main(void){ is not a valid signature for main.
It should be:

int main(void){

Then:

float vett[10];
vett[10] = 0;

This is not valid. Array indices start at 0, so index 10 is out of bounds, as it would require an array with size 11.

Also, as your average function takes as first argument a float array, you'll need to pass it this way:

average(vett,media);

Using:

average(&vett[10],media);

Will pass a pointer to the data located right after the array, so obviously you'll get junk values.

Solution 2

• Don't use func(void) in C++, it's some sort of deprecated. Use func() instead.
• If you have float vett[10];, then vett[10] is invalid. You must use only vett[0 .. 9]
• float average(float array[10], float &average)
  actually means
  float average(float *array, float &average).
  (Second form is more common way to declare pointer arguments.)
  If you want to call it with vett as argument, just use average(vett, media);
  
  Names of an arrays, when used as pointers, are automatically converted to pointer to first element of an array. So here
  average(vett, media);
vett means &vett[0].

Solution 3

The number in the square brackets has two meanings:

• In a declaration like float vett[10]; it is the size of the array
• When not in a declaration, like average(&vett[10],media); it means the eleventh element of the array.

average(&vett[10],media); is passing the address of the eleventh element to the function. The function interprets it as the beginning of the array, which is wrong and causes undefined behaviour when the elements outside of the array are accessed.

Because you want to pass the whole array, you should use

average(vett,media);

Comments

[typ1232]

vett[10] = 0; This accesses memory outside of the array bounds and doesn't do what you want. Do you want to zero initialize the array? Use float vett[10] = {0} in the declaration.

[crashmstr]

And I just saw the same kind of question a few days ago.

[PaulMcKenzie]

return array[10]; return average; What are you expecting to accomplish by using return twice?

[noelicus]

No. 10 means the 11th element in the array.

[noelicus]

Nope. That was a pretty good reason! Have removed it after your edit.

[alain]

Thanks, and I fully agree. I'm not so good at playing "The fastest gun in the west"...