使用由函数传递的数组。

Question

// This program demonstrates an array being passed to a function 

#include <iostream>
using namespace std;

void showValues(int [], int) ; //Function prototype.

int main()
{
    const int ARRAY_SIZE = 8;
    int number [ARRAY_SIZE] = {5, 10, 15, 20, 25, 30, 35, 40};

    showValues (number, ARRAY_SIZE);
    return 0;
}   
    //Definition of function showValue.
    //This function accpets an array to integers and 
    //the array's size as its arguments. The contents. 
    //of the array are displayed. 

    void showValues (int nums[], int size)
    {
        for (int index = 0; index < size; index++)
            cout << nums [index] << " ";
            cout << endl;

    }

Using C++, Studying arrays, the program works fine, however; I am not understanding the for loop on the bottom where it says "index < size" Where is "size" getting its value from that the for loop knows when to stop looping?

Answer 1

size is passed as a parameter to the function showValues .

That function is called from main() , passing ARRAY_SIZE as that parameter.

ARRAY_SIZE is defined as the size of the array.

Actually, you could write

int number [] = {5, 10, 15, 20, 25, 30, 35, 40};

since the compiler can figure out the array size from the initialisation array. The function call can be refactored to

showValues(number, sizeof(number) / sizeof(int));

sizeof(number) / sizeof(int) is an idiomatic way of evaluating the number of elements in an array. Some folk prefer sizeof(number) / sizeof(number[0]) since then the code is not sensitive to the type of the array. You can then remove ARRAY_SIZE entirely.

Answer 2

The showValues (number, ARRAY_SIZE); call

passes the array as well as its size to void showValues (int nums[], int size)

Therefore size in the for loop gets the value 8

Answer 3

 void showValues (int nums[], int size)

The second parameter to the function showValues becomes the size in your for loop.

const int ARRAY_SIZE = 8;

showValues (number, ARRAY_SIZE);