print array member function

This is a discussion on print array member function within the C++ Programming forums, part of the General Programming Boards category; I am trying to write a simple function that will print my array of clocks. I have a function called ...

  1. #1
    Registered User brianptodd's Avatar
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    print array member function

    I am trying to write a simple function that will print my array of clocks. I have a function called display_time() for the clocks:

    Code:
    void clock::display_time()
    {
    	cout << "Current time is: " << _hours << ":" << setfill ('0') << setw(2) << _mins << " ";
    	if (_morn)
    		cout << "AM" << endl;
    	else
    		cout << "PM" << endl;
    }
    And my print function is:

    Code:
    void clock::print_clocks(clock& array)
    {
    	for (int i = 0; i < array._used; i ++)
    	{
    		cout << array[i].display_time() << endl;
    	}
    }
    The error I'm getting at compile is:

    "no match for clock&[int&]" at the "cout << array[i].... line."

    What does this mean?

    Brian
    "In theory, there is no difference between theory and practice. But, in practice, there is."
    - Jan L.A. van de Snepscheut

  2. #2
    Cat
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    void clock::print_clocks(clock* array)
    You ever try a pink golf ball, Wally? Why, the wind shear on a pink ball alone can take the head clean off a 90 pound midget at 300 yards.

  3. #3
    Registered User brianptodd's Avatar
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    That made things much worse. Should I be passing a pointer to the function?
    "In theory, there is no difference between theory and practice. But, in practice, there is."
    - Jan L.A. van de Snepscheut

  4. #4
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    You haven't shown us the entire declaration of your clock class, so we have no idea what your intent is there. Does the clock class store an array of all your clocks with _used counting the number and operator[] providing access? If not, then you'll have to pass an array (which decays to a pointer) to the function.

    By the way, if print_clocks() is receiving an array of clocks and not using or modifying any member data, then there is really no reason for it to be a member function.

  5. #5
    Cat
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    You are trying to pass an array? Arrays are always passed as pointers to their first member.

    You will also need to pass in the array size, as in this example:

    Code:
    #include <iostream>
    
    void PrintArray(int * array, int count){
    	for (int i = 0; i < count; i++)
    		std::cout << "[" << array[i] << "]";
    	std::cout << std::endl;
    }
    
    int main(){
    	int myArray[5] = {6,2,1,3,4};
    	PrintArray(myArray,5);
    	int * myArray2 = new int[6];
    	for (int i = 0 ;i < 6; i++)
    		myArray2[i] = i * 2 + 1;
    	PrintArray(myArray2,6);
    	delete[] myArray2;
    
    }
    You ever try a pink golf ball, Wally? Why, the wind shear on a pink ball alone can take the head clean off a 90 pound midget at 300 yards.

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